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Example 5. Show that the system of functions
$$
\left\{\begin{array}{l}
x_{1}(t)=C_{1} e^{-t}+C_{2} e^{3 t} \\
x_{2}(t)=2 C_{1} e^{-t}-2 C_{2} e^{3 t}
\end{array}\right.
$$
is the general solution of the system of equations
$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=x_{1}-x_{2} \\
\frac{d x_{2}}{d t}=x_{2}-4 x_{... | Solution. In this example, the domain $D$ is
$$
-\infty<t<+\infty, \quad-\infty<x_{1}, x_{2}<+\infty .
$$
Substituting the functions $x_{1}(t)$ and $x_{2}(t)$ from (6) into the system of equations (7), we obtain identities in $t$ that hold for any values of the constants $C_{1}, C_{2}$. Thus, the condition 1) definin... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,512 |
Example 6. Given the general solution (6) of system (7), find the particular solution of this system that satisfies the initial conditions $x_{1}(0)=0$, $x_{2}(0)=-4$. | Solution. The problem reduces to finding such values of the constants $C_{1}$ and $C_{2}$ that the following relations hold:
$$
0=C_{1}+C_{2}, \quad-4=2 C_{1}-2 C_{2}
$$
Solving this system, we find $C_{1}=-1, C_{2}=1$. The desired particular solution is
$$
x_{1}(t)=-e^{-t}+e^{3 t}, \quad x_{2}(t)=-2 e^{-t}-2 e^{3 t... | x_{1}()=-e^{-}+e^{3},\quadx_{2}()=-2e^{-}-2e^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,513 |
Example 7. Solve the system of equations
$$
\frac{d x}{d t}=y, \quad \frac{d y}{d t}=-x
$$
with initial conditions
$$
x(0)=x_{0}, \quad y(0)=y_{0}
$$ | Solution. Differentiating the first equation of system (10) once with respect to $t$ and substituting $\frac{d y}{d t}=-x$ into the resulting equation, we reduce system (10) to a single second-order equation $\frac{d^{2} x}{d t^{2}}+x=0$, the general solution of which is
$$
x=C_{1} \cos t+C_{2} \sin t
$$
Since $y=\fr... | x^{2}+y^{2}=R^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,514 |
Example 8. Show that the function
$$
\psi\left(t, x_{1}, x_{2}\right)=\frac{x_{2}}{t}-x_{1}
$$
defined in the domain $D: t \neq 0, -\infty < x_{1}, x_{2} < +\infty$, is an integral of the system of equations
$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=\frac{x_{1}}{t} \\
\frac{d x_{2}}{d t}=x_{1}+\frac{x_{2}}{t}
\... | Solution. Substituting (19) into (17), we get
$$
\psi\left(t, x_{1}, x_{2}\right)=\psi\left(t, C_{1} t, C_{1} t^{2}+C_{2} t\right)=\frac{C_{1} t^{2}+C_{2} t}{t}-C_{1} t=C_{2}
$$
in the domain $D$. Therefore, the function (17) is an integral of the system of equations (18) in the domain $D$, which means the first inte... | \frac{x_{2}}{}-x_{1}=C | Calculus | proof | Yes | Yes | olympiads | false | 34,515 |
Example 9. Show that the function
$$
\psi\left(t, x_{1}, x_{2}\right)=\operatorname{arctg} \frac{x_{1}}{x_{2}}-t
$$
is an integral of the system of equations
$$
\frac{d x_{1}}{d t}=\frac{x_{1}^{2}}{x_{2}}, \quad \frac{d x_{2}}{d t}=-\frac{x_{2}^{2}}{x_{1}}
$$[^5] | Solution. In this case
$$
f_{1}\left(t, x_{1}, x_{2}\right)=\frac{x_{1}^{2}}{x_{2}}, \quad f_{2}\left(t, x_{1}, x_{2}\right)=-\frac{x_{2}^{2}}{x_{1}}
$$
We find the partial derivatives of the function $\psi\left(t, x_{1}, x_{2}\right)$. We have
$$
\frac{\partial \psi}{\partial t}=-1, \quad \frac{\partial \psi}{\part... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,516 |
Example 1. Integrate the system of equations
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=y+1 \\
\frac{d y}{d t}=x+1
\end{array}\right.
$$ | Solution. From the first equation of the system (3), we find $y=\frac{d x}{d t}-1$, then
$$
\frac{d y}{d t}=\frac{d^{2} x}{d t^{2}}
$$
Substituting (4) into the second equation of the system (3), we obtain a second-order linear differential equation with constant coefficients
$$
\frac{d^{2} x}{d t^{2}}-x-1=0
$$
The... | C_{1}e^{}+C_{2}e^{-}-1,\quadC_{1}e^{}-C_{2}e^{-}-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,517 |
Example 2. Solve the Cauchy problem for the system
$$
\begin{gathered}
\left\{\begin{array}{l}
\frac{d x}{d t}=3 x+8 y \\
\frac{d y}{d t}=-x-3 y
\end{array}\right. \\
x(0)=6, \quad y(0)=-2
\end{gathered}
$$ | Solution. From the second equation of the system (7), we find
$$
x=-3 y-\frac{d y}{d t}
$$
from which
$$
\frac{d x}{d t}=-3 \frac{d y}{d t}-\frac{d^{2} y}{d t^{2}}
$$
Substituting (9) and (10) into the first equation of the system (7), we obtain the equation $\frac{d^{2} y}{d t^{2}}-y=0$, the general solution of wh... | 4e^{}+2e^{-},\quad-e^{}-e^{-} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,518 |
Example 3. Solve the system of equations
$$
\left\{\begin{array}{l}
t \frac{d x}{d t}=-x+y t \\
t^{2} \frac{d y}{d t}=-2 x+y t
\end{array}\right.
$$ | Solution. From the first equation of the system, we find
$$
y=\frac{x}{t}+\frac{d x}{d t}, \quad \text { so that } \quad \frac{d y}{d t}=-\frac{x}{t^{2}}+\frac{1}{t} \cdot \frac{d x}{d t}+\frac{d^{2} x}{d t^{2}} \text {. }
$$
Substituting these expressions for $y$ and $\frac{d y}{d t}$ into the second equation, we ge... | C_{1}+C_{2},\quad\frac{C_{1}}{}+2C_{2},\quad\neq0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,519 |
Example 1. Solve the system
$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=2\left(x_{1}^{2}+x_{2}^{2}\right) t \\
\frac{d x_{2}}{d t}=4 x_{1} x_{2} t
\end{array}\right.
$$ | Solution. Adding the two equations term by term, we get
$$
\frac{d\left(x_{1}+x_{2}\right)}{d t}=\left(x_{1}+x_{2}\right)^{2} 2 t
$$
from which
$$
-\frac{1}{x_{1}+x_{2}}=t^{2}-C_{1}, \text { or } \frac{1}{x_{1}+x_{2}}+t^{2}=C_{1} \text {. }
$$
Subtracting the two equations term by term, we get
$$
\frac{d\left(x_{1... | x_{1}=\frac{C_{1}+C_{2}-2^{2}}{2(C_{1}-^{2})(C_{2}-^{2})},\quadx_{2}=\frac{C_{2}-C_{1}}{2(C_{1}-^{2})(C_{2}-^{2})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,520 |
Example 2. Solve the system
$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=\frac{x_{1}-x_{2}}{x_{3}-t} \\
\frac{d x_{2}}{d t}=\frac{x_{1}-x_{2}}{x_{3}-t} \\
\frac{d x_{3}}{d t}=x_{1}-x_{2}+1
\end{array}\right.
$$ | Solution. Subtracting the second equation from the first term by term, we get $\frac{d\left(x_{1}-x_{2}\right)}{d t}=0$, from which the first integral of the system (4)
$$
x_{1}-x_{2}=C_{1}
$$
Substituting (5) into the second and third equations of the system (4), we obtain a system with two unknown functions
$$
\le... | x_{1}=\ln|C_{1}+C_{2}|+C_{1}+C_{3},\quadx_{2}=\ln|C_{1}+C_{2}|+C_{3},\quadx_{3}=(C_{1}+1)+C_{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,521 |
Example 3. Find a particular solution of the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=1-\frac{1}{y} \\
\frac{d y}{d t}=\frac{1}{x-t}
\end{array}\right.
$$
satisfying the initial conditions $\left.x\right|_{t=0}=1,\left.y\right|_{t=0}=1$. | Solution. Let's write the given system in the form
$$
\left\{\begin{array} { c }
{ y ( \frac { d x } { d t } - 1 ) = - 1 , } \\
{ ( x - t ) \frac { d y } { d t } = 1 , }
\end{array} \quad \text { or } \quad \left\{\begin{array}{c}
y \frac{d(x-t)}{d t}=-1 \\
(x-t) \frac{d y}{d t}=1
\end{array}\right.\right.
$$
Adding... | +e^{-},\quade^{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,522 |
Example 4 (Decomposition of a Substance). Substance $A$ decomposes into two substances $X$ and $Y$ with the rate of formation of each being proportional to the amount of undecomposed substance. Find the law of change of the quantities $x$ and $y$ of substances $X$ and $Y$ as functions of time $t$, if at $t=0$ we have $... | Solution. At time $t$, the amount of undecomposed substance $A$ is $a-x-y$. By the condition of the problem, we have
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=k_{1}(a-x-y) \\
\frac{d y}{d t}=k_{2}(a-x-y)
\end{array}\right.
$$
Dividing the second equation by the first, we get
$$
\frac{d y}{d t}=\frac{k_{2}}{k_{1}}, ... | \frac{}{4}(1-2^{-}),\quad\frac{3}{4}(1-2^{-}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,523 |
Example 5 (equilibrium of gases in communicating vessels). Let there be vessels of volumes $V_{1}$ and $V_{2}$ respectively, filled with gas. The gas pressure at the initial moment of time is $P_{1}$ in the first vessel and $P_{2}$ - in the second. The vessels are connected by a tube, through which the gas flows from o... | Solution. Let $a$ be the amount of gas that flows in a unit of time when the pressure difference is one unit. Then, during the time $d t$, the amount of gas that flows from one vessel to another is $a\left(p_{1}^{2}-p_{2}^{2}\right) d t$. This amount is equal to the loss of gas in one vessel and the gain in the other o... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,524 |
Example 6. Find the general solution of the system of equations
$$
\frac{d t}{2 x}=\frac{d x}{-\ln t}=\frac{d y}{\ln t-2 x}
$$ | Solution. The first integrable combination is $\frac{d t}{2 x}=-\frac{d x}{\ln t}$. Separating variables and integrating, we find the first integral
$$
t(\ln t-1)+x^{2}=C_{1} .
$$
The second integrable combination is obtained using the derivatives of the proportion (19). For this, we add the numerators and denominato... | \\sqrt{C_{1}+(\ln-1)},\quadC_{2}-\\sqrt{C_{1}+(\ln-1)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,525 |
Example 7. Solve the system of equations
$$
\frac{d t}{4 y-5 x}=\frac{d x}{5 t-3 y}=\frac{d y}{3 x-4 t}
$$ | Solution. Multiplying the numerators and denominators of the fractions in system (23) by $3, 4, 5$ respectively and adding the numerators and denominators, we obtain, by (19),
$$
\frac{3 d t}{12 y-15 x}=\frac{4 d x}{20 t-12 y}=\frac{5 d y}{15 x-20 t}=\frac{3 d t+4 d x+5 d y}{0}
$$
(here $\lambda_{1}=3, \lambda_{2}=4,... | 3+4x+5C_{1},\quad^{2}+x^{2}+y^{2}=C_{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,526 |
Example 1. Solve the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=3 x-y+z \\
\frac{d y}{d t}=-x+5 y-z \\
\frac{d z}{d t}=x-y+3 z
\end{array}\right.
$$ | Solution. We form the characteristic equation
$$
\left|\begin{array}{ccc}
3-r & -1 & 1 \\
-1 & 5-r & -1 \\
1 & -1 & 3-r
\end{array}\right|=0
$$
or $r^{3}-11 r^{2}+36 r-36=0$.
The roots $r_{1}=2, r_{2}=3, r_{3}=6$ correspond to the numbers
$$
\begin{array}{lll}
\lambda_{1}=1, & \mu_{1}=0, & \nu_{1}=-1 \\
\lambda_{2}... | \begin{aligned}&C_{1}e^{2}+C_{2}e^{3}+C_{3}e^{6}\\&C_{2}e^{3}-2C_{3}e^{6}\\&-C_{1}e^{2}+C_{2}e^{3}+C_{3}e^{6}\end{aligned} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,527 |
Example 2. Solve the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=x-5 y \\
\frac{d y}{d t}=2 x-y
\end{array}\right.
$$ | Solution. Let's write down the system for determining $\lambda$ and $\mu$:
$$
\left\{\begin{array}{l}
(1-r) \lambda-5 \mu=0 \\
2 \lambda-(1+r) \mu=0
\end{array}\right.
$$
The characteristic equation
$$
\left|\begin{array}{cc}
1-r & -5 \\
2 & -1-r
\end{array}\right|=0
$$
has roots $r_{1}=3 i, r_{2}=-3 i$. Substituti... | \begin{aligned}&5C_{1}\cos3+5C_{2}\sin3\\&C_{1}(\cos3+3\sin3)+C_{2}(\sin3-3\cos3)\end{aligned} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,528 |
Example 3. Solve the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=2 x+y \\
\frac{d y}{d t}=4 y-x
\end{array}\right.
$$ | Solution. The characteristic equation
$$
\left|\begin{array}{cc}
2-r & 1 \\
-1 & 4-r
\end{array}\right|=0, \text { or } r^{2}-6 r+9=0
$$
has a multiple root $r_{1}=r_{2}=3$.
The solution should be sought in the form
$$
x=\left(\lambda_{1}+\mu_{1} t\right) e^{3 t}, \quad y=\left(\lambda_{2}+\mu_{2} t\right) e^{3 t}
... | (C_{1}+C_{2})e^{3},\quad(C_{1}+C_{2}+C_{2})e^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,529 |
Example 4. Solve the Cauchy problem for the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=8 y \\
\frac{d y}{d t}=-2 z \\
\frac{d z}{d t}=2 x+8 y-2 z
\end{array}\right.
$$
with initial conditions $x(0)=-4, y(0)=0, z(0)=1$. | Solution. Characteristic equation
$$
\left|\begin{array}{rrc}
-r & 8 & 0 \\
0 & -r & -2 \\
2 & 8 & -2-r
\end{array}\right|=0, \text { or }(r+2)\left(r^{2}+16\right)=0 \text {. }
$$
The roots of equation (17): $r_{1}=-2, r_{2}=4 i, r_{3}=-4 i$. The real root $r_{1}=-2$ corresponds to the solution
$$
x_{1}=\lambda_{1}... | -4e^{-2}-2\sin4,\quade^{-2}-\cos4,\quade^{-2}-2\sin4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,530 |
Example 1. Solve the system by the method of variation of parameters
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=-2 x-4 y+1+4 t \\
\frac{d y}{d t}=-x+y+\frac{3}{2} t^{2}
\end{array}\right.
$$ | Solution. First, we solve the corresponding homogeneous system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=-2 x-4 y \\
\frac{d y}{d t}=-x+y
\end{array}\right.
$$
From the second equation of the system (8) we have
$$
x=y-\frac{d y}{d t}, \quad \text { so } \frac{d x}{d t}=\frac{d y}{d t}-\frac{d^{2} y}{d t^{2}}
$$
Su... | -C_{1}e^{2}+4C_{2}e^{-3}++^{2},\quadC_{1}e^{2}+C_{2}e^{-3}-\frac{1}{2}+^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,531 |
Example 2. Find the general solution of the non-homogeneous system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=x-2 y+e^{t} \\
\frac{d y}{d t}=x+4 y+e^{2 t}
\end{array}\right.
$$ | Solution. First, we find the general solution of the corresponding homogeneous system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=x-2 y \\
\frac{d y}{d t}=x+4 y
\end{array}\right.
$$
The characteristic equation is
$$
\left|\begin{array}{cc}
1-\lambda & -2 \\
1 & 4-\lambda
\end{array}\right|=0, \text { or } \quad \lam... | \begin{aligned}&2C_{1}e^{2}+C_{2}e^{3}-\frac{3}{2}e^{}+2e^{2}\\&-C_{1}e^{2}-C_{2}e^{3}+\frac{1}{2}e^{}+(+1)e^{2}\end{aligned} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,532 |
Example 3. Solve the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=x+2 y \\
\frac{d y}{d t}=x-5 \sin t
\end{array}\right.
$$ | Solution. The characteristic equation
$$
\left|\begin{array}{cc}
1-\lambda & 2 \\
1 & -\lambda
\end{array}\right|=0, \quad \text { or } \quad \lambda^{2}-\lambda-2=0
$$
The roots of the characteristic equation are $\lambda_{1}=-1, \lambda_{2}=2$. The general solution of the corresponding homogeneous system:
$$
\tild... | \begin{aligned}&C_{1}e^{-}+2C_{2}e^{2}-\cos+3\sin\\&-C_{1}e^{-}+C_{2}e^{2}+2\cos-\sin\end{aligned} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,533 |
Example 4. Solve the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=x+2 y+16 t e^{t} \\
\frac{d y}{d t}=2 x-2 y
\end{array}\right.
$$ | Solution. The characteristic equation
$$
\left|\begin{array}{cc}
1-\lambda & 2 \\
2 & -2-\lambda
\end{array}\right|=0, \text { or } \quad \lambda^{2}+\lambda-6=0
$$
The roots of the characteristic equation are $\lambda_{1}=2, \lambda_{2}=-3$. The general solution of the homogeneous system corresponding to system (15)... | \begin{aligned}&2C_{1}e^{2}+C_{2}e^{-3}-(12+13)e^{}\\&C_{1}e^{2}-2C_{2}e^{-3}-(8+6)e^{}\end{aligned} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,534 |
Example 5. Solve the system by D'Alembert's method
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=5 x+4 y+e^{t} \\
\frac{d y}{d t}=4 x+5 y+1
\end{array}\right.
$$ | Solution. We choose $\lambda$ according to formula (19): $4+5 \lambda=\lambda(5+4 \lambda)$, from which $\lambda_{1,2}= \pm 1$. Then, by formula (20) for the case $\lambda=1$, we have
$$
\begin{aligned}
& x+y=e^{(s+4 \cdot 1) t}\left\{C_{1}+\int\left(e^{t}+1\right) e^{-(s+4 \cdot 1) t} d t\right\}= \\
& \quad=e^{9 t}\... | (x+y+\frac{1}{8}e^{}+\frac{1}{9})e^{-9}=C_{1},\quad(x-y-e^{}-1)e^{-}=C_{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,535 |
Example 1. Find the original $f(t)$, if
$$
F(p)=\frac{p+2}{(p+1)(p-2)\left(p^{2}+4\right)}
$$ | Solution. First method. Represent $F(p)$ as a sum of simple fractions
$$
\frac{p+2}{(p+1)(p-2)\left(p^{2}+4\right)}=\frac{A}{p+1}+\frac{B}{p-2}+\frac{C_{p}+D}{p^{2}+4}
$$
and find the undetermined coefficients $A, B, C, D$. We have
$$
p+2=A(p-2)\left(p^{2}+4\right)+B(p+1)\left(p^{2}+4\right)+\left(C_{p}+D\right)(p+1... | f()=-\frac{1}{15}e^{-}+\frac{1}{6}e^{2}-\frac{1}{10}\cos2-\frac{1}{5}\sin2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,536 |
Example 2. Find the original $f(t)$, if $F(p)=\frac{p+2}{p^{3}(p-1)^{2}}$. | Solution. The fraction $F(p)$ has a pole $p_{1}=0$ of multiplicity $n_{1}=3$ and a pole $p_{2}=1$ of multiplicity $n_{2}=2$. Using formula (16), we obtain the original
$$
\begin{aligned}
f(t)= & \frac{1}{2} \lim _{p \rightarrow 0} \frac{d^{2}}{d p^{2}}\left[\frac{p+2}{p^{3}(p-1)^{2}} p^{3} e^{p t}\right]+\lim _{p \rig... | 8+5+^{2}+(3-8)e^{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,537 |
Example 3. Solve the equation
$$
\begin{array}{r}
x^{\prime}+x=1 \\
x(0)=1
\end{array}
$$ | Solution. Let $x(t) \risingdotseq X(p)$, then by the rule of differentiation of the original we have
$$
x^{\prime}(t) \risingdotseq p X(p)-x(0)=p X(p)-1 .
$$
It is known that $1 \risingdotseq \frac{1}{p}$, therefore, transitioning from the given problem (21)-(22) to the operator equation, we will have
$$
p X(p)-1+X(... | x()\equiv1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,538 |
Example 4. Solve the equation $x^{\prime \prime}-5 x^{\prime}+4 x=4, x(0)=0, x^{\prime}(0)=2$. | Solution. Since $4 \risingdotseq \frac{4}{p}$ and by the condition $x_{0}=x(0)=0, x_{1}=x^{\prime}(0)=2$, the operator equation will have the form $\left(p^{2}-5 p+4\right) X(p)=\frac{4}{p}+2$. From this, we find the operator solution
$$
X(p)=\frac{2 p+4}{p\left(p^{2}-5 p+4\right)}
$$
We decompose the right-hand side... | x()=1-2e^{}+e^{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,539 |
Example 6. Find the solution to the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=-7 x+y+5 \\
\frac{d y}{d t}=-2 x-5 y-37 t
\end{array}\right.
$$
satisfying the initial condition $x(0)=0, y(0)=0$. | Solution. Since $5 \risingdotseq \frac{5}{p}, -37 t = -\frac{37}{p^{2}}$ and $x_{0}=y_{0}=0$, the operator system will have the form
$$
\left\{\begin{array}{l}
p X(p) = -7 X(p) + Y(p) + \frac{5}{p} \\
p Y(p) = -2 X(p) - 5 Y(p) + \frac{37}{p^{2}}
\end{array}\right.
$$
Solving this, we get
$$
X(p) = \frac{5 p^{2} + 25... | x()=1--e^{-6}\cos,\quady()=1-7+e^{-6}\cos+e^{-6}\sin | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,540 |
Example 1. Based on the definition of Lyapunov stability, investigate the stability of the solution to the equation
$$
\frac{d x}{d t}=1+t-x
$$
satisfying the initial condition
$$
x(0)=0 \text {. }
$$ | Solution. Equation (5) is a linear inhomogeneous equation. Its general solution is $x(t)=C e^{-t}+t$. The initial condition $x(0)=0$ is satisfied by the solution
$$
\varphi(t)=t \quad t
$$
of equation (5). The initial condition $x(0)=x_{0}$ is satisfied by the solution
$$
x(t)=x_{0} e^{-t}+t
$$
Consider the differe... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,541 |
Example 2. Investigate the stability of the solution of the equation
$$
\frac{d x}{d t}=\sin ^{2} x
$$ | Solution. It has obvious solutions
$$
x=k \pi, \quad k=0, \pm 1, \pm 2, \ldots
$$
Integrating equation (8):
$$
\operatorname{ctg} x=C-t, \quad \text { or } \quad \operatorname{ctg} x=\operatorname{ctg} x_{0}-t,
$$
from which
$$
x=\operatorname{arcctg}\left(\operatorname{ctg} x_{0}-t\right), \quad x \neq k \pi
$$
... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,542 |
Example 3. Based on the definition of Lyapunov stability, show that the solution of the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=-y \\
\frac{d y}{d t}=x
\end{array}\right.
$$
satisfying the initial conditions $x(0)=0, y(0)=0$, is stable. | Solution. The solution of system (11) that satisfies the given initial conditions is $x(t) \equiv 0, y(t) \equiv 0$. Any solution of this system that satisfies the conditions $x(0)=x_{0}, y(0)=y_{0}$ has the form
$$
x(t)=x_{0} \cos t - y_{0} \sin t, \quad y(t)=x_{0} \sin t + y_{0} \cos t
$$
Take any $\varepsilon > 0$... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,543 |
Example 4. Investigate the stability of the solution of the nonlinear equation
$$
\frac{d x}{d t}=1-x^{2}(t)
$$ | Solution. It has obvious solutions $\varphi(t)=-1$ and $\varphi(t)=1$.
The solution $\varphi(t)=-1$ of this equation is unstable, while the solution $\varphi(t)=1$ is asymptotically stable. Indeed, as $t \rightarrow+\infty$, all solutions of equation (14)
$$
x(t)=\frac{\left(1+x_{0}\right) e^{2\left(t-t_{0}\right)}-\... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,544 |
Example 1. Determine the character of the equilibrium point (0,0) of the system
\[
\left\{\begin{array}{l}
\frac{d x}{d t}=5 x-y \\
\frac{d y}{d t}=2 x+y
\end{array}\right.
\] | Solution. We form the characteristic equation
$$
\left|\begin{array}{cc}
5-\lambda & -1 \\
2 & 1-\lambda
\end{array}\right|=0, \quad \text { or } \quad \lambda^{2}-6 \lambda+7=0
$$
Its roots $\lambda_{1}=3+\sqrt{2}>0, \lambda_{2}=3-\sqrt{2}>0$ are real, distinct, and positive. Therefore, the equilibrium point (0,0) i... | unstablenode | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,545 |
Example 2. Investigate the equation of elastic oscillations
$$
\frac{d^{2} x}{d t^{2}}+2 \alpha \frac{d x}{d t}+\beta^{2} x=0
$$
taking into account friction and resistance of the medium (for $\alpha>0$). | Solution. We transition from equation (3) to an equivalent system of equations
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=y \\
\frac{d y}{d t}=-2 \alpha y-\beta^{2} x
\end{array}\right.
$$
To determine the nature of the equilibrium point (0,0) of system (4), we form the characteristic equation
$$
\left|\begin{array}... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,546 |
Example 3. Will the equilibrium point $(0,0)$ of the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=-x+z \\
\frac{d y}{d t}=-2 y-z \\
\frac{d z}{d t}=y-z
\end{array}\right.
$$
be stable? | Solution. We form the characteristic equation
$$
\left|\begin{array}{ccc}
-1-\lambda & 0 & 1 \\
0 & -2-\lambda & -1 \\
0 & 1 & -1-\lambda
\end{array}\right|=0
$$
or $(1+\lambda)\left(\lambda^{2}+3 \lambda+3\right)=0$. The roots of this equation are $\lambda_{1}=-1, \lambda_{2,3}=-\frac{3}{2} \pm i \frac{\sqrt{3}}{2}$... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,547 |
Example 1. Investigate the stability of the equilibrium point of the system
\[
\left\{\begin{array}{l}
\frac{d x}{d t}=y \\
\frac{d y}{d t}=-x
\end{array}\right.
\] | Solution. We choose the function $V(x, y)$ to be $V=x^{2}+y^{2}$. This function is positive definite. The derivative of the function $V$ according to system (4) is
$$
\frac{d V}{d t}=2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=2 x y-2 x y \equiv 0
$$
From Theorem 1, it follows that the equilibrium point $O(0,0)$ of syste... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,548 |
Example 2. Investigate the stability of the equilibrium point of the system
\[
\left\{\begin{array}{l}
\frac{d x}{d t}=y-x^{3} \\
\frac{d y}{d t}=-x-3 y^{3}
\end{array}\right.
\] | Solution. Taking again $V(x, y)=x^{2}+y^{2}$, we find
$$
\frac{d V}{d t}=2 x\left(y-x^{3}\right)+2 y\left(-x-3 y^{3}\right)=-2\left(x^{4}+3 y^{4}\right)
$$
Thus, $\frac{d V}{d t}$ is a definitively negative function. By Theorem 2, the equilibrium point $O(0,0)$ of system (5) is asymptotically stable.
There is no gen... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,549 |
Example 3. Investigate the stability of the trivial solution $x \equiv 0, y \equiv 0$ of the system using the Lyapunov function
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=-x-2 y+x^{2} y^{2} \\
\frac{d y}{d t}=x-\frac{y}{2}-\frac{x^{3} y}{2}
\end{array}\right.
$$ | Solution. We will look for a Lyapunov function in the form \( V = a x^{2} + b y^{2} \), where \( a > 0, b > 0 \) are arbitrary parameters. We have
\[
\begin{aligned}
\frac{d V}{d t} = \frac{\partial V}{\partial x} \cdot \frac{d x}{d t} + & \frac{\partial V}{\partial y} \frac{d y}{d t} = 2 a x \left( -x - 2 y + x^{2} y... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,550 |
Example 4. Investigate the stability of the equilibrium point $x=0, y=0$ of the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=x \\
\frac{d y}{d t}=-y
\end{array}\right.
$$ | Solution. Let's take the function $v(x, y)=x^{2}-y^{2}$. Then
$$
\frac{d v}{d t}=\frac{\partial v}{\partial x} \frac{d x}{d t}+\frac{\partial v}{\partial y} \frac{d y}{d t}=2 x^{2}+2 y^{2}
$$
is a function that is definitely positive. Since arbitrarily close to the origin there are points where $v>0$ (for example, $v... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,551 |
Example 5. Investigate the stability of the equilibrium point $x=0, y=0$ of the system
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=y^{3}+x^{5} \\
\frac{d y}{d t}=x^{3}+y^{5}
\end{array}\right.
$$ | Solution. The function $v=x^{4}-y^{4}$ satisfies the conditions of Chetaev's theorem:
1) $v>0$ when $|x|>|y|$;
2) $\frac{d v}{d t}=4\left(x^{8}-y^{8}\right)$ is definitively positive in the region $|x|>|y|$.
Therefore, the equilibrium point $x=0, y=0$ is unstable.
## Problems for Independent Solution
Investigate th... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,552 |
Example 1. Investigate the stability of the equilibrium point $x=0, y=0$ of the system by the first approximation
$$
\left\{\begin{array}{l}
\dot{x}=2 x+y-5 y^{2} \\
\dot{y}=3 x^{6}+y+\frac{x^{3}}{2}
\end{array}\left(\dot{x}=\frac{d x}{d t}, \quad \dot{y}=\frac{d y}{d t}\right)\right.
$$ | Solution. Systems of the first approximation
$$
\left\{\begin{array}{l}
\dot{x}=2 x+y \\
\dot{y}=3 x+y
\end{array}\right.
$$
Nonlinear terms satisfy the necessary conditions: their order is greater than or equal to two. We form the characteristic equation for system (6):
$$
\left|\begin{array}{cc}
2-\lambda & 1 \\
3... | \lambda_{1}=\frac{3+\sqrt{13}}{2},\lambda_{2}=\frac{3-\sqrt{13}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,553 |
Example 2. Investigate the stability of the equilibrium point $x \equiv 0, y \equiv 0$ by the first approximation for the systems
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=y-x^{3} \\
\frac{d y}{d t}=-x-y^{3}
\end{array}\right.
$$
and
$$
\left\{\begin{array}{l}
\frac{d x}{d t}=y+x^{3} \\
\frac{d y}{d t}=-x+y^{3}
\en... | Solution. The equilibrium point $x \equiv 0, y \equiv 0$ of system (8) is asymptotically stable because for this system the function $v=x^{2}+y^{2}$ satisfies all the conditions of Lyapunov's theorem on asymptotic stability. In particular,
$$
\frac{d v}{d t}=2 x\left(y-x^{3}\right)+2 y\left(-x-y^{3}\right)=-2\left(x^{... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,554 |
Example 1. In the square $Q\left\{-\frac{1}{2} \leqslant x \leqslant \frac{1}{2} ;-\frac{1}{2} \leqslant y \leqslant \frac{1}{2}\right\}$, find an approximate solution to the equation
$$
y'=\sin (x y)
$$
satisfying the initial condition
$$
\left.y\right|_{x=0}=0.1
$$
and estimate the error. | Solution. Replace equation (4) with the equation
\[
\begin{aligned}
y^{\prime} & =x y \\
\left.y\right|_{x=0} & =0.1
\end{aligned}
\]
Equation (6) with the initial condition (7) has the solution \( y=0.1 \cdot e^{x^{2} / 2} \), which for all \( x \in \left[-\frac{1}{2} ; \frac{1}{2}\right] \) does not leave the main ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,556 |
Example 1. Investigate the stability of the zero solution of the equation
$$
y^{\text {IV }}+5 y^{\prime \prime \prime}+13 y^{\prime \prime}+19 y^{\prime}+10 y=0
$$ | Solution. We form the characteristic equation
$$
f(\lambda) \equiv \lambda^{4}+5 \lambda^{3}+13 \lambda^{2}+19 \lambda+10=0
$$
Here $a_{0}=1, a_{1}=5, a_{2}=13, a_{3}=19, a_{4}=10$. We list the Hurwitz diagonal minors
$$
\begin{gathered}
\Delta_{4}=\left|\begin{array}{rrrr}
5 & 1 & 0 & 0 \\
19 & 13 & 5 & 1 \\
0 & 10... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,557 |
Example 2. Investigate the stability of the solution of the degenerate equation for the equation
$$
\varepsilon \frac{d x}{d t}=x\left(e^{x}-2\right)
$$ | Solution. The degenerate equation $x\left(e^{x}-2\right)=0$ has two solutions
$$
\text { 1) } x=0, \quad 2) x=\ln 2 \text {. }
$$
We have
$$
\left.\frac{\partial f(t, x)}{\partial x}\right|_{x=0}=\left.\left(e^{x}-2+x e^{x}\right)\right|_{x=0}=-1
$$
so the solution $x=0$ is stable;
$$
\left.\frac{\partial f(t, x)}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,559 |
Example 3. Investigate the stability of the solution of the degenerate equation corresponding to the equation $\varepsilon \frac{d x}{d t}=(x-t)^{2}$. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,560 | |
## 1.1.
First, let's familiarize ourselves with the theoretical concepts and knowledge related to the divisibility of polynomials.
The canonical form of a polynomial with variable \( x \) is its representation as a sum of monomials in decreasing (or increasing) powers of \( x \).
The general form of a polynomial \( ... | $\triangle$ Consider the following division scheme:
$$
\begin{array}{cc}
2 x^{4}-x^{3}-3 x^{2}+x+1 & x^{2}+x+2 \\
\hline 2 x^{4}+2 x^{3}+4 x^{2} & 2 x^{2}-3 x-4 \\
-3 x^{3}-7 x^{2}+x+1 & \\
-3 x^{3}-3 x^{2}-6 x & \\
-\frac{-4 x^{2}+7 x+1}{-4 x^{2}-4 x-8} &
\end{array}
$$
$$
11 x+9
$$
This should be understood as fol... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,561 |
1. Find all integer roots of the polynomial:
$$
f(x)=4 x^{4}-16 x^{3}+11 x^{2}+4 x-3
$$ | $\triangle$ By Theorem 6, all integer roots of the given polynomial are among the divisors of the constant term, i.e., among the numbers $\pm 1, \pm 3$.
Let's check if $x=1$ is a root of the polynomial. We will have:
$$
f(1)=4-16+11+4-3=0
$$
Therefore, $x=1$ is a root of the polynomial.
For $x=-1, x=3$, and $x=-3$,... | 1,3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,562 |
3. Find all rational roots of the polynomials:
a) $x^{4}-3 x^{3}-8 x^{2}+12 x+16$
b) $8 x^{3}-20 x^{2}-2 x+5$
c) $4 x^{4}-16 x^{3}+11 x^{2}+4 x-3$. | $\triangle$ a) The polynomial is monic, so by Theorem 7, all its rational roots are integers. We find them using the familiar method; we get the roots $-1, 2, -2, 4$.
b) Consider the equation
$$
8 x^{3}-20 x^{2}-2 x+5=0
$$
We make the polynomial on its left side monic. For this, let $2 x=y$. We will have:
$$
y^{3}-... | )-1,\2,4;b)\1/2,5/2;)\1/2,1,3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,563 |
8. Find the remainders of dividing the polynomial
$$
f(x)=x^{243}+x^{81}+x^{27}+x^{9}+x^{3}+1
$$
by a) $x-1$; b) $x^{2}-1$. | $\triangle$ a) By the Remainder Theorem, the remainder from dividing the given polynomial by $x-1$ is $f(1)=6$.
b) It is possible to divide the given polynomial by $x^{2}-1$ using long division, but this is a lengthy method. It is simpler to isolate terms from the sum that are divisible by $x^{2}-1$:
$$
f(x)=\left(x^... | )6;b)5x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,564 |
10. Find all values of parameters $a$ and $b$, for which the polynomial
$$
f(x)=x^{5}-3 x^{4}+a x^{3}+b x^{2}-5 x-5
$$
is divisible by $x^{2}-1$ without a remainder. | $\triangle$ Polynomial $f(x)$, by the condition, is divisible by $x+1$ and $x-1$. Then, based on Theorem 4, the values $x=-1$ and $x=1$ are its roots. We obtain the system of equations:
$$
\left\{\begin{array}{r}
-1-3-a+b+5-5=0 \\
1-3+a+b-5-5=0
\end{array}\right.
$$
Solving it, we find $a$ and $b$.
Answer: $a=4, b=8... | =4,b=8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,565 |
12. Find all values of the coefficients $a, b$, and $c$ of the polynomial
$$
f(x)=x^{4}-x^{3}+a x^{2}+b x+c
$$
if it is known that it is divisible by the polynomial $\varphi(x)=x^{3}-2 x^{2}-5 x+6$ without a remainder. | $\triangle$ Let's divide the first polynomial by the second with a remainder:
$$
\begin{gathered}
x^{4}-x^{3}+a x^{2}+b x+c=\left(x^{3}-2 x^{2}-5 x+6\right)(x+1)+ \\
+(a+7) x^{2}+(b-1) x+(c-6)
\end{gathered}
$$
The remainder must be identically equal to zero:
$$
(a+7) x^{2}+(b-1) x+(c-6) \equiv 0
$$
Since two polyn... | =-7,b=1,=6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,566 |
21. Prove that among the roots of the polynomial
$$
f(x)=x^{3}+a x^{2}+b x+c
$$
there are two opposite numbers if and only if $a b=c$, where $b \leq 0$. | $\triangle$ In the proof, we will consider two cases.
1) Let the equality $a b=c$ hold, where $b \leq 0$. Then the equation $f(x)=0$ takes the form:
$$
x^{3}+a x^{2}+b x+a b=0, \quad(x+a)\left(x^{2}+b\right)=0
$$
The roots of the last equation are $-x_{1}=-a, x_{2,3}= \pm \sqrt{-b}$.
It turns out that among the roo... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,567 |
26*. Does there exist an integer $a$ such that the polynomial $f(x)=x^{13}+x+90$ is divisible by the trinomial $x^{2}-x+a$? | $\triangle$ Since the polynomial $f(x)$ is divisible by $x^{2}-x+a$, then for any integer $x$ the number $f(x)$ is divisible by the number $x^{2}-x+a$ (in the sense of divisibility of integers):
$$
\left(x^{13}+x+90\right) \vdots\left(x^{2}-x+a\right)
$$
Let's substitute $x=1$, and then $x=0$:
$$
92 \vdots a, 90 \vd... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,568 |
28*. Prove that if a polynomial $p(x)$ with integer coefficients takes the same value, equal to 1, at three different integer values of the argument, then it does not have integer roots. | $\triangle$ Polynomial $p(x)-1$ by condition has three distinct roots. Let's denote them as $a, b$ and $c$. Then
$$
p(x)-1=(x-a)(x-b)(x-c) \varphi(x)
$$
where $\varphi(x)$ is a polynomial with integer coefficients.
Suppose that the polynomial $p(x)$ has an integer root $m$. Set $x=m$ in the above identity:
$$
-1=(m... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,569 |
30*. Prove that if a polynomial $p(x)$ has a root $x=a$ of multiplicity $k (k>1)$, then its derivative has a root $x=a$ of multiplicity $k-1$.
31${}^{\circ}$. Prove that if $f(x)$ is a polynomial with rational coefficients and
$$
f(a+b \sqrt{c})=P+Q \sqrt{c},
$$
where the numbers $a, b, c, P, Q$ are rational, $c>0$,... | $\triangle$ It is sufficient to prove this statement for the power $x^{n} (n \in N)$. We will use the method of mathematical induction.
(For the method of mathematical induction, see [15] or [19].)
For $n=1$, our statement is, of course, true. It is also true for $n=2$:
$$
\begin{aligned}
& (a+b \sqrt{c})^{2}=\left(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,570 |
34. The number $2+\sqrt{3}$ is a root of the polynomial
$$
f(x)=x^{3}+4 x^{2}+a x+b(a \in Q, b \in Q)
$$
Find the other roots. | $\triangle$ First, let's find the values of the parameters $a$ and $b$.
Based on the statement of the reference problem 32, if $x_{1}=2+\sqrt{3}$ is a root of the given polynomial, then $x_{2}=2-\sqrt{3}$ is also its root. Therefore, the polynomial $f(x)$ is divisible by the expression
$\left(x-x_{1}\right)\left(x-x_... | 2-\sqrt{3},-8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,571 |
41*. Factor the polynomial with real coefficients:
$$
x\left(y^{2}-z^{2}\right)+y\left(z^{2}-x^{2}\right)+z\left(x^{2}-y^{2}\right)
$$ | $\triangle$ From this sum, we can factor out the term $y-z$. We will have:
$$
\begin{aligned}
& x\left(y^{2}-z^{2}\right)+y\left(z^{2}-x^{2}\right)+z\left(x^{2}-y^{2}\right)=x(y+z)(y-z)+y z^{2}-y x^{2}+z x^{2}-z y^{2}= \\
& =(y-z)(x y+x z)-y z(y-z)-x^{2}(y-z)=(y-z)\left(x y+x z-y z-x^{2}\right)= \\
& =(y-z)(x(z-x)-y(z... | (x-y)(y-z)(z-x) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,572 |
47. Factor the polynomial $x^{4}+4 y^{4}$ with real coefficients. | ## $\triangle$ We have:
$x^{4}+4 y^{4}=\left(x^{4}+4 y^{4}+4 x^{2} y^{2}\right)-4 x^{2} y^{2}=\left(x^{2}+2 y^{2}\right)^{2}-4 x^{2} y^{2}=$ $=\left(x^{2}+2 x y+2 y^{2}\right)\left(x^{2}-2 x y+2 y^{2}\right)$
Answer: $\left(x^{2}+2 x y+2 y^{2}\right)\left(x^{2}-2 x y+2 y^{2}\right)$. | (x^{2}+2xy+2y^{2})(x^{2}-2xy+2y^{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,573 |
49*. Factor the polynomial with integer coefficients:
$$
x^{5}+x+1
$$ | $\triangle$ Subtract and add $x^{2}$. We get:
$x^{5}+x+1=\left(x^{5}-x^{2}\right)+\left(x^{2}+x+1\right)=x^{2}\left(x^{3}-1\right)+\left(x^{2}+x+1\right)=$ $=x^{2}(x-1)\left(x^{2}+x+1\right)+\left(x^{2}+x+1\right)=\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\right)$.
Answer: $\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\ri... | (x^{2}+x+1)(x^{3}-x^{2}+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,574 |
51. Find all integers $a$ for which the polynomial
$$
y=x^{4}-3 x^{3}+a x^{2}-9 x-2
$$
can be factored into the product of two quadratic trinomials with integer coefficients. | $\triangle$ The coefficients of $x^{2}$ in both quadratic trinomials can be set to 1. As for the constant terms, since their product is -2, they are either 2 and -1, or 1 and -2.
Consider the two cases.
1) Suppose the constant terms of the quadratic trinomials are 2 and -1. We get:
$$
x^{4}-3 x^{3}+a x^{2}-9 x-2=\le... | -3,-11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,575 |
54. Construct some polynomial $p(x)$ of degree greater than 2, which takes the values 1, 2, and 3 at $x=1, 2, 3$ respectively.
$55^{\circ}$. Prove that if $p(x)$ is a polynomial with integer coefficients, then for any distinct integers $b$ and $c$, the difference $p(b)-p(c)$ is divisible by $b-c$.
This problem is fou... | $\triangle$ Consider the polynomial
$$
p(x)=a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n-1} x+a_{n},
$$
where all coefficients $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ are integers, and $a_{0} \neq 0$. We compute $p(b)$ and $p(c)$:
$$
\begin{aligned}
& p(b)=a_{0} b^{n}+a_{1} b^{n-1}+a_{2} b^{n-2}+\cdots+a_{n-1} b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,576 |
56. Does there exist a polynomial of the third degree
$$
p(x)=a x^{3}+b x^{2}+c x+d
$$
where $a, b, c, d$ are integers, $a \neq 0$, such that $p(15)=3, p(21)=12$? | $\triangle$ Here
$$
p(21)-p(15)=12-3=9, \quad 21-15=6 \text{.}
$$
It turned out that the difference $p(21)-p(15)$ is not divisible by $21-15$, which contradicts the statement of problem 55.
Answer: does not exist. | doesnotexist | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,577 |
57. Does there exist a polynomial $p(x)$ with integer coefficients such that:
a) $p(1)=19, p(19)=99$;
b) $p(0)=19, p(1)=99, p(2)=1999$?
$58^{\circ}$. Prove that the sum of all coefficients of any polynomial $p(x)$ is equal to its value at $x=1$. | $\triangle$ Indeed, let us set $x=1$ in the equality. We get:
$$
p(x)=a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n}
$$
$$
p(1)=a_{0}+a_{1}+a_{2}+\cdots+a_{n}
$$
Our statement is proved. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,578 |
65. Find all integers $a$ such that the quadratic trinomial
$$
y=(x-a)(x-6)+1
$$
can be represented as the product $(x+b)(x+c)$, where $b$ and $c$ are integers. | $\triangle$ Let's substitute $x=-b$ into the identity
$$
(x-a)(x-6)+1=(x+b)(x+c)
$$
We get:
$$
(-b-a)(-b-6)+1=0, \quad(b+a)(b+6)=-1
$$
From this, either $b+a=1, b+6=-1$, i.e., $b=-7, a=8$, or $b+a=-1$, $b+6=1$, i.e., $b=-5, a=4$.
The problem could also be solved in another way: in the identity written above, expan... | 4or8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,579 |
68*. All coefficients of the polynomial $p(x)$ are natural numbers less than 10. It is known that $p(10)=1248$. Find this polynomial. | $\triangle$ Let's write $p(x)$ in the form
$$
p(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}
$$
Let's set $x=10$ here. Then
$$
a_{0} 10^{n}+a_{1} 10^{n-1}+\cdots+a_{n-1} 10+a_{n}=1248
$$
From this,
$$
a_{0} 10^{n}+a_{1} 10^{n-1}+\cdots+a_{n-1} \cdot 10=1240+\left(8-a_{n}\right)
$$
Therefore, the difference... | p(x)=x^{3}+2x^{2}+4x+8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,580 |
69. Can the polynomial $y=5 x^{20}+2$ be represented as the sum of squares of two polynomials with integer coefficients? | $\triangle$ Suppose it is possible:
$$
5 x^{20}+2=P^{2}(x)+Q^{2}(x)
$$
where $P(x)$ and $Q(x)$ are polynomials with integer coefficients. Let's set $x=1$ here. Then
$$
P^{2}(x)+Q^{2}(x)=7
$$
But the sum of two perfect squares cannot be equal to 7. We have reached a contradiction.
Answer: It is not possible. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,581 |
71*. Do there exist polynomials $P(x, y)$ and $Q(x, y)$ in two variables $x$ and $y$ such that for all real $x$ and $y$ the equality
$$
(x+y) P(x, y)+(2 x-y-3) Q(x, y)=x^{2}+y^{2} ?
$$ | $\triangle$ Let's choose values of $x$ and $y$ such that
$$
x+y=0, \quad 2 x-y-3=0
$$
Solving this system of equations, we will have: $x=1, y=-1$.
Now let's substitute $x=1, y=-1$ into the identity. We get the incorrect equality $0=2$. Therefore, such polynomials do not exist.
Answer: do not exist. | donotexist | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,582 |
73*. Polynomial
$$
y=x^{3}-18 x^{2}+a x+1784
$$
takes on three consecutive integer values at three consecutive integer points (in the same order). Find these values. | $\triangle$ Let's denote three consecutive integer values of the argument as $k-1$, $k$, and $k+1$, and the corresponding values of the polynomial as $n-1$, $n$, and $n+1$. We obtain the system of equations:
$$
\left\{\begin{array}{l}
(k-1)^{3}-18(k-1)^{2}+a(k-1)+1784=n-1 \\
k^{3}-18 k^{2}+a k+1784=n \\
(k+1)^{3}+18(k... | 1999,2000,2001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,583 |
75*. Prove that if a polynomial $p(x)$ with integer coefficients takes odd values at $x=0$ and $x=1$, then it does not have integer roots. | ## $\triangle$ Let
$$
p(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}
$$
where the coefficients $a_{0}, a_{1}, \ldots, a_{n}$ are integers, and $a_{0} \neq 0$.
Since $p(0)=a_{n}$ and $p(0)$ is odd by condition, $a_{n}$ is odd. If the polynomial has an integer root, then it is a divisor of the constant term $a_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,584 |
77*. Quadratic trinomial
$$
p(x)=a x^{2}+b x+c
$$
is such that the equation $p(x)=x$ has no real roots. Prove that then the equation $p(p(x))=x$ also has no real roots. | $\triangle$ Since the equation $p(x)=x$ has no real roots, the quadratic trinomial $p(x)-x$ maintains a constant sign for all $x$, which coincides with the sign of the coefficient $a$ of its leading term.
Let, for example, $a>0$. Then for all $x$
$$
p(x)-x>0, \quad p(x)>x
$$
Therefore, for all $x$
$$
p(p(x))>p(x)
$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,585 |
81. Calculate the product:
$$
P=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \cdots\left(1-\frac{1}{n^{2}}\right)
$$ | $\triangle$ Let's simplify this expression:
$$
P=\frac{2^{2}-1}{2^{2}} \cdot \frac{3^{2}-1}{3^{2}} \cdot \frac{4^{2}-1}{4^{2}} \cdots \cdots \cdot \frac{n^{2}-1}{n^{2}}=\frac{1 \cdot 3}{2^{2}} \cdot \frac{2 \cdot 4}{3^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdots \cdot \frac{(n-1)(n+1)}{n^{2}}
$$
The obtained product can... | \frac{n+1}{2n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,586 |
86. Calculate the sum:
$$
S=\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}(x \neq \pm 1)
$$ | $\triangle$ Let's use the identity $\frac{1}{1-x}+\frac{1}{1+x}=\frac{2}{1-x^{2}}$. We get:
$$
S=\frac{2}{1-x^{2}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}=\frac{4}{1-x^{4}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}=\frac{8}{1-x^{8}}+\frac{8}{1+x^{8}}=\frac{16}{1-x^{16}}
$$
Answer: $16 /\left(1-x^{16}\right)$. | \frac{16}{1-x^{16}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,587 |
88. Calculate the sum:
$$
S=\frac{1}{(a-b)(a-c)}+\frac{1}{(b-a)(b-c)}+\frac{1}{(c-a)(c-b)}
$$ | $$
\begin{aligned}
& S=\frac{1}{(a-b)(b-c)(c-a)}(-(b-c)-(c-a)-(a-b))= \\
& =\frac{-1}{(a-b)(b-c)(c-a)}(b-c+c-a+a-b)=0
\end{aligned}
$$
Answer: 0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,588 |
93. Calculate the sum:
$$
S=1-2+3-4+\cdots+(-1)^{n+1} \cdot n
$$ | $\triangle$ Clearly, the answer depends on whether $n$ is even or odd. Therefore, we will consider two cases.
1) Let $n$ be even. Then:
$$
S=(1-2)+(3-4)+(5-6)+\cdots+((n-1)-n)=-1-1-\ldots-1=-\frac{n}{2}
$$
2) Let $n$ be odd. We will have:
$$
S=(1-2+3-4+\cdots-(n-1))+n=-\frac{n-1}{2}+n=\frac{n+1}{2}
$$
Answer: if $... | if\n\is\even,\then\S=-\frac{n}{2};\if\n\is\odd,\then\S=\frac{n+1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,589 |
97. Prove the identity:
$$
n\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}\right)=\frac{n-1}{1}+\frac{n-2}{2}+\frac{n-3}{3}+\cdots+\frac{1}{n-1}(n>1) .
$$ | $\triangle$ Transform the left side of the equality:
$$
\begin{aligned}
& n\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}\right)=\frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\cdots+\frac{n}{n-1}+1= \\
& =\frac{(n-2)+2}{2}+\frac{(n-3)+3}{3}+\frac{(n-4)+4}{4}+\cdots+\frac{(n-1)+1}{n-1}+1= \\
& =\frac{n-2}{2}+1+\fra... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,590 |
101. Prove that for any $x$ the following equality holds:
$$
a^{2} \cdot \frac{(x-b)(x-c)}{(a-b)(a-c)}+b^{2} \cdot \frac{(x-c)(x-a)}{(b-c)(b-a)}+c^{2} \cdot \frac{(x-a)(x-b)}{(c-a)(c-b)}=x^{2}
$$ | $\triangle$ Instead of a straightforward transformation of the left side of this equality, let's try the following approach. For $x=a, x=b$, and $x=c$, we obtain the valid equalities
$$
a^{2}=a^{2}, \quad b^{2}=b^{2}, \quad c^{2}=c^{2}
$$
But the left side is a quadratic trinomial. If we move $x^{2}$ from the right s... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,591 |
103*. Prove the identity:
$$
\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2 n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}(n \in N)
$$ | $\triangle$ Transform the left side of this equality by adding and subtracting the sum $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$
$\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2 n}=$
$=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}\right)-\left(1+\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,592 |
106. Calculate the sum:
$$
S=\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{99 \cdot 100}
$$ | $\triangle$ Let's represent the first term of the sum as $1-\frac{1}{2}$, the second term as $\frac{1}{2}-\frac{1}{3}$, the third term as $\frac{1}{3}-\frac{1}{4}$, and generally, the $k$-th term as
$$
\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}
$$
We get:
$$
S=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}... | 0.99 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,593 |
108. Calculate the sum:
$$
S=\frac{3}{1^{2} \cdot 2^{2}}+\frac{5}{2^{2} \cdot 3^{2}}+\frac{7}{3^{2} \cdot 4^{2}}+\cdots+\frac{2 n+1}{n^{2}(n+1)^{2}}
$$ | $\triangle$ Let's represent the $k$-th term of this sum ( $k=1,2, \ldots, n$ ) in the following form:
$$
\frac{2 k+1}{k^{2}(k+1)^{2}}=\frac{(k+1)^{2}-k^{2}}{k^{2}(k+1)^{2}}=\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}
$$
Now we will have:
$\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\fra... | \frac{n(n+2)}{(n+1)^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,594 |
114. Calculate the sum:
$$
x=\sqrt[3]{9+4 \sqrt{5}}+\sqrt[3]{9-4 \sqrt{5}}
$$ | $$
\triangle$ Let's raise the equality to the third power, using the formula
$$
(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)
$$
We get:
$$
\begin{aligned}
& x^{3}=(9+4 \sqrt{5})+(9-4 \sqrt{5})+3 \sqrt[3]{81-80} \cdot(\sqrt[3]{9+4 \sqrt{5}}+\sqrt[3]{9-4 \sqrt{5}}) \\
& x^{3}=18+3 x, \quad x^{3}-3 x-18=0
\end{aligned}
$$
The las... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,595 |
115. Calculate the values of the expressions:
a) $\sqrt[3]{26+15 \sqrt{3}}+\sqrt[3]{26-15 \sqrt{3}}$;
b) $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$.
$2^{*}$
35
$\mathbf{1 1 6}^{\circ}$. Prove the identity:
$$
\sqrt{a \pm \sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^{2}-b}}{2}}\left(a>0, b>0, ... | $\triangle$ Let's prove the equality for the plus sign in both parts of the equation. We will square it in this case:
$a+\sqrt{b}=\frac{a+\sqrt{a^{2}}-b}{2}+\frac{a-\sqrt{a^{2}}-b}{2}+2 \sqrt{\frac{a^{2}-\left(a^{2}-b\right)}{4}}, \quad a+\sqrt{b}=a+\sqrt{b}$.
The proven formula is called the formula of the complex s... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,596 |
117. Simplify the expression:
$$
\sqrt{2+\sqrt{3}}
$$ | $\triangle$ Using the formula for the compound square radical, we get:
$$
\sqrt{2+\sqrt{3}}=\sqrt{\frac{2+\sqrt{4-3}}{2}}+\sqrt{\frac{2-\sqrt{4-3}}{2}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}=\frac{\sqrt{2}(\sqrt{3}+1)}{2}
$$
Answer: $\sqrt{2}(\sqrt{3}+1) / 2$ | \frac{\sqrt{2}(\sqrt{3}+1)}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,597 |
123. Rationalize the denominator of the fraction:
$$
\frac{1}{\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{4}+\sqrt[5]{2}+1}
$$ | $\triangle$ Let's use the identity
$$
a^{5}-b^{5}=(a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)
$$
(see § 1, item 1.1). Let $a=\sqrt[5]{2}, b=1$; then $a^{5}-b^{5}=2-1=1$. Therefore, the conjugate factor for the denominator of the given fraction is $\sqrt[5]{2}-1$.
Answer: $\sqrt[5]{2}-1$. | \sqrt[5]{2}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,598 |
128. Knowing that $x+\frac{1}{x}=4$, calculate $x^{2}+\frac{1}{x^{2}}$. | $\triangle$ It seems that the equation $x+\frac{1}{x}=4$ should be transformed into a quadratic equation, find $x$ from the quadratic equation, and substitute the obtained value of $x$ into the sum $x^{2}+\frac{1}{x^{2}}$. However, this path is long, especially since the quadratic equation has two roots.
It is simpler... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,599 |
151. Prove that if $\frac{x^{2}-y z}{x(1-y z)}=\frac{y^{2}-x z}{y(1-x z)}(x \neq y)$, then
$$
x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
$$ | $\triangle$ Transform the original equality by eliminating the denominators of the fractions $\left(x^{2}-y z\right) y(1-x z)=\left(y^{2}-x z\right) x(1-y z)$, $x^{2} y-y^{2} z-x^{3} y z+x y^{2} z^{2}=x y^{2}-x^{2} z-x y^{3} z+x^{2} y z^{2}$, $x y(x-y)+z\left(x^{2}-y^{2}\right)-x y z\left(x^{2}-y^{2}\right)-x y z^{2}(x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,600 |
154. Eliminate $x$ and $y$ from the equations:
$$
a=x+y, \quad b=x^{3}+y^{3}, \quad c=x^{5}+y^{5}
$$ | $\triangle$ Let's raise the first equality to the third power and subtract the second equality:
$$
a^{3}-b=(x+y)^{3}-x^{3}-y^{3}=3 x y(x+y)=3 a x y, \quad x y=\frac{a^{3}-b}{3 a}
$$
Let's raise the first equality to the second power and multiply by the second equality:
$a^{2} b=(x+\mathrm{y})^{2}\left(x^{3}+y^{3}\ri... | 5b(^{3}+b)=(^{5}+9) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,601 |
159. Prove that if
$$
\frac{a-b}{1+a b}+\frac{b-c}{1+b c}+\frac{c-a}{1+c a}=0
$$
then at least one of the fractions in the left part of the equality is zero. | $\triangle$ The problem reduces to proving that among the differences
$$
a-b, \quad b-c, \quad c-a
$$
there is one equal to zero.
Let's eliminate the denominators in the original equation:
$$
(a-b)(1+b c)(1+c a)+(b-c)(1+a b)(1+c a)+(c-a)(1+a b)(1+b c)=0 .
$$
In the left part of this equation, factor out $a-b$ by e... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,602 |
161. It is known that $a b(a+b)=1$. Prove that
$$
\frac{a}{a^{3}+a+1}=\frac{b}{b^{3}+b+1}
$$ | $$
\begin{gathered}
\frac{a}{a^{3}+a+1}=\frac{a}{a^{3}+a+a^{2} b+a b^{2}}=\frac{1}{a^{2}+1+a b+b^{2}}= \\
=\frac{b}{a^{2} b+b+a b^{2}+b^{3}}=\frac{b}{b^{3}+b+1} \cdot \mathbf{\Delta}
\end{gathered}
$$
Let's represent the condition in the following form: $1=a^{2} b+a b^{2}$. Now we get:
$$
\begin{gathered}
\frac{a}{a^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,603 |
164. Prove that if
$$
|a-b| \geq|c|, \quad|b-c| \geq|a|, \quad|c-a| \geq|b|
$$
then one of the numbers $a, b$, and $c$ is equal to the sum of the other two. | $\triangle$ Note that the condition here is expressed by inequalities.
Let's square each of these inequalities (thus obtaining equivalent inequalities), move all terms to the left sides of the inequalities, and factor the differences of squares:
$(a-b)^{2}-c^{2} \geq 0, \quad(b-c)^{2}-a^{2} \geq 0, \quad(c-a)^{2}-b^{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,604 |
165. It is known that
$$
S=\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0
$$
Prove that then
$$
S^{\prime}=\frac{a}{(b-c)^{2}}+\frac{b}{(c-a)^{2}}+\frac{c}{(a-b)^{2}}=0
$$ | $\triangle$ Multiply the original equality respectively by
$$
\frac{1}{b-c}, \quad \frac{1}{c-a}, \frac{1}{a-b}
$$
We will have:
$$
\begin{aligned}
& \frac{a}{(b-c)^{2}}+\frac{b}{(b-c)(c-a)}+\frac{c}{(b-c)(a-b)}=0 \\
& \frac{a}{(c-a)(b-c)}+\frac{b}{(c-a)^{2}}+\frac{c}{(c-a)(a-b)}=0 \\
& \frac{a}{(a-b)(b-c)}+\frac{b}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,605 |
167. Prove that the sequence $\left(a_{n}\right)$, defined by the relations
$$
a_{1}=2, \quad a_{n+1}=-\frac{1}{a_{n}}
$$
is periodic. | $\triangle$ Let's compute the first few terms of this sequence:
$$
a_{1}=2, \quad a_{2}=-\frac{1}{2}, \quad a_{3}=2, \quad a_{4}=-\frac{1}{2}
$$
Obviously, this pattern will continue, since
$$
a_{n+2}=-\frac{1}{a_{n+1}}=-1:\left(-\frac{1}{a_{n}}\right)=a_{n}
$$
Therefore, this sequence is periodic with period $p=2$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,606 |
172. Prove that the numbers formed by the last two digits of the sequence $a_{n}=6^{n}$ form a periodic sequence. Find the period of such a sequence. | $\triangle$ That such a sequence is periodic is almost obvious: after all, the numbers formed by the last two digits of the power $6^{n}$ do not exceed 10: from 06 to 96. To find the period of the sequence, we will compute its first few terms:
$$
06,36,16,96,76,56,36,16
$$
It turned out that $a_{7}=a_{2}$, which mean... | 5 | Number Theory | proof | Yes | Yes | olympiads | false | 34,607 |
175. The sequence $\left(x_{n}\right)$ is defined by the conditions:
$$
x_{1}=a, \quad x_{n+1}=x_{n}^{2}
$$
Find all values of $a$ for which it is periodic. | $\triangle$ From the condition, it follows that for $n \geq 2$, $x_{n}$ is non-negative.
We will investigate the sequence for increasing and decreasing behavior. For this, we will determine the sign of the difference $x_{n+1}-x_{n}$. We will have:
$$
x_{n+1}-x_{n}=x_{n}^{2}-x_{n}=x_{n}\left(x_{n}-1\right)
$$
From th... | 0,\1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,608 |
179. The sequence $\left(a_{n}\right)$ satisfies the conditions:
$$
a_{1}=1, \quad a_{2}=5, \quad a_{n+1}=2 a_{n}-a_{n-1}(n>1)
$$
Find the formula for the general term of the sequence. | $\triangle$ The last condition can be represented in the following form:
$$
a_{n+1}-a_{n}=a_{n}-a_{n-1}
$$
This equality means that the sequence $\left(a_{n}\right)$ is an arithmetic progression. Its first term $a_{1}=1$, and the common difference
$$
d=a_{2}-a_{1}=5-1=4
$$
Using the formula for the $n$-th term of a... | a_{n}=4n-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,609 |
181. The sequence $\left(x_{n}\right)$ satisfies the conditions:
$$
x_{1}=1 ; \quad x_{n+1}=3 x_{n}+2^{n}
$$
Prove that for any natural $n$, $x_{n}=3^{n}-2^{n}$. | $\triangle$ Apply the method of mathematical induction.
1) For $n=1$ this formula is valid, since $x_{1}=3-2=1$.
2) Assume that it is valid for some $n=k$ :
$$
x_{k}=3^{k}-2^{k}
$$
Then for $n=k+1$ we get:
$$
x_{k+1}=3 x_{k}+2^{k}=3 \cdot\left(3^{k}-2^{k}\right)+2^{k}=3^{k+1}-3 \cdot 2^{k}+2^{k}=3^{k+1}-2^{k+1}
$$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,610 |
184. The sequence $\left(a_{n}\right)$ satisfies the conditions:
$$
a_{1}=3, \quad a_{2}=9, \quad a_{n+1}=4 a_{n}-3 a_{n-1}(n>1)
$$
Find the formula for the general term of the sequence. | $\triangle$ Let's compute the first few terms of the sequence, starting from $a_{3}$:
$a_{3}=4 a_{2}-3 a_{1}=4 \cdot 9-3 \cdot 3=27=3^{3}$,
$a_{4}=4 a_{3}-3 a_{2}=4 \cdot 27-3 \cdot 9=81=3^{4}$
$a_{5}=4 a_{4}-3 a_{3}=4 \cdot 81-3 \cdot 27=243=3^{5}$.
It seems that for any natural number $n$, $a_{n}=3^{n}$. Prove th... | a_{n}=3^{n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,611 |
186. The sequence $\left(a_{n}\right)$ is defined by the conditions:
$$
a_{1}=0, \quad a_{n+1}=a_{n}+n
$$
Find the formula for the general term of the sequence. | $\triangle$ Let's represent the second condition in the following form:
$$
a_{k+1}=a_{k}+k, \quad a_{k+1}-a_{k}=k
$$
Let's set $k=1,2,3, \ldots, n-1$ in the last equality. We will obtain a series of equalities, which we will then add term by term:
$$
\begin{aligned}
& a_{2}-a_{1}=1 \\
& a_{3}-a_{2}=2, \\
& +a_{4}-a_... | a_{n}=\frac{n(n-1)}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,612 |
189*. The sequence $\left(a_{n}\right)$ satisfies the conditions:
$$
a_{1}=0, \quad a_{n+1}=\frac{n}{n+1}\left(a_{n}+1\right)
$$
Find the formula for $a_{n}$. | $\triangle$ Transform the second condition as follows:
$$
a_{n+1}+1=\frac{n}{n+1}\left(a_{n}+1\right)+1, \quad(n+1)\left(a_{n+1}+1\right)=n\left(a_{n}+1\right)+(n+1)
$$
The last equality suggests introducing a new sequence $b_{n}=n\left(a_{n}+1\right)$. It satisfies the initial condition $b_{1}=1$ and the recurrence ... | a_{n}=\frac{n-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,613 |
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