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852. In the definition: "a locus of points is the set of all points possessing a certain property," explain the importance of the word "all." Illustrate with examples. | 852. The word "all" is of essential importance in this definition because it indicates that outside the line (or surface, if we are talking about space), all points of which have the required property, there are no other points that also possess it. For example, the geometric locus of points in a plane from which a giv... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 40,451 |
858. What does the expression "necessary and sufficient condition" mean? Explain the meaning of the words "necessary" and "sufficient" separately and in combination. | 858. If from position $\boldsymbol{A}$ follows position $\boldsymbol{B}$, then it is said that $\boldsymbol{A}$ implies $\boldsymbol{B}$, or: $\boldsymbol{A}$ is a sufficient condition for $\boldsymbol{B}$; it can also be said that $\boldsymbol{B}$ is a consequence of $\boldsymbol{A}$, or $\boldsymbol{B}$ is a necessar... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 40,452 |
862. What does the method of proof by moving from $n$ to $n+1$ (so-called complete induction) consist of? Explain with examples.
The method of proof by moving from $n$ to $n+1$, also known as complete induction or mathematical induction, is a powerful technique used to prove statements that are indexed by natural numb... | 862. The essence of this method is as follows. 1) The validity of the theorem that needs to be proven for any natural number \( n \) is first established directly (verified) for some small value of the number \( n \), for example, for \( n=2 \). 2) Then we prove that since the theorem is valid for a certain value \( n=... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,453 |
863. What does the method of proof by reduction to absurdity (reductio ad absurdum) consist of? Outline the general line of reasoning; provide an example.
The method of proof by reduction to absurdity, also known as proof by contradiction or reductio ad absurdum, involves assuming the opposite of what you want to prov... | 863. The method of proof by reduction to absurdity (proof "by contradiction") proceeds as follows. Assume for a moment that what needs to be proven is false, that the opposite statement is true; starting from this assumption, we make a series of arguments that lead us to a clearly false, "absurd" conclusion, i.e., a co... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,454 |
Example 2. Are the equation
$$
\left(x^{2}+x+1\right)(3 x+4)(-7 x+2)(2 x-\sqrt{5})(-12 x-16)=0
$$
and the system of equations
$$
3 x+4=0, \quad-7 x+2=0, \quad 2 x-\sqrt{\overline{5}}=0, \quad-12 x-16=0
$$
equivalent on the set of all real numbers? | Solved. Since
$$
x^{2}+x+1=(x+1 / 2)^{2}+3 / 4
$$
then for any $x$ the inequality $x^{2}+x+1>0$ holds. Thus, the given equation is equivalent to the equation
$$
(3 x+4)(-7 x+2)(2 x-\sqrt{5})(-12 x-16)=0
$$
Any root of this equation makes at least one of the polynomials zero, i.e., it is a root of at least one of th... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,456 |
Example 3. Are the equation
$$
3 \log _{3}|-x|=\log _{3} x^{2}
$$
and the system of equations
$$
x+1=0, \quad x-1=0, \quad x=0
$$
equivalent on the domain of definition of the given equation? | Solution. The domain of admissible values for this equation is the set $R \backslash\{0\}$. On this set, the given system has two roots: $x_{1}=-1$ and $x_{2}=1$. Both these numbers, and only they, are the roots of the equation. Therefore, the given equation and the system of equations are equivalent on the domain of t... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,457 |
Example 10. Are the equations
$$
x+7+\frac{10}{2 x-1}=8-x+\frac{10}{2 x-1} \quad \text{ and } \quad x+7=8-x
$$
equivalent? | Solution. The second equation is obtained from the first equation by adding the same expression $-\frac{10}{2 x-1}$ to both of its parts, which is not defined at $x=1 / 2$. This means that the number $1 / 2$ cannot be a root of the first equation, but it can be a root of the second. It is easy to check that the number ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,458 |
Example 11. Are the equations
$$
\frac{2(x-10)}{x^{2}-13 x+30}=1 \quad \text { and } \quad x^{2}-15 x+50=0
$$
equivalent? | Solution. We will solve the first equation. By getting rid of the denominator, i.e., multiplying both sides of the original equation by the expression $x^{2}-13 x+30$, we obtain the equation
$$
2 x-20=x^{2}-13 x+30
$$
The set of all roots of this equation consists of two numbers: $x_{1}=10$ and $x_{2}=5$. As a result... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,459 |
Example 12. Are the equations
$$
\sqrt{x^{2}+x-5}=\sqrt{x-1} \quad \text { and } \quad x^{2}+x-5=x-1
$$
equivalent? | Solution. The set of all roots of the second equation consists of two numbers: $x_{1}=2$ and $x_{2}=-2$. Checking shows that the number -2 does not belong to the domain of the first equation and therefore cannot be its root; hence, these equations are not equivalent.
## CH. 1. EQUIVALENT EQUATIONS AND INEQUALITIES
Th... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,460 |
Example 13. Are the equations
$$
2 \sqrt{x+5}=x+2 \quad \text { and } \quad 4(x+5)=(x+2)^{2}
$$
equivalent? | Solution. The set of all roots of the second equation consists of two numbers: 4 and -4. However, the number -4 is not a root of the first equation; therefore, these equations are not equivalent. At the same time, the number -4 satisfies the condition $x \geqslant-5$, i.e., it belongs to the domain of the first equatio... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,461 |
Example 14. Are the equations
$$
\lg \left(x^{2}-4\right)=\lg (4 x-7) \quad \text { and } \quad x^{2}-4=4 x-7
$$
equivalent? | Solution. The set of all roots of the second equation consists of the numbers $x_{1}=3$ and $x_{2}=1$. However, the number 1 is not a root of the first equation, and therefore these equations are not equivalent.
This example shows that the transition from the equation $\log _{a} f(x)=$ $=\log _{a} g(x)$ (where $a>0, a... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,462 |
Example 15. Given two equations:
$$
\frac{f_{1}(x)}{f_{2}(x)}=\frac{f_{3}(x)}{f_{4}(x)} \quad \text { and } \quad \frac{f_{1}(x)}{f_{2}(x)}=\frac{f_{1}(x)+f_{3}(x)}{f_{2}(x)+f_{4}(x)}
$$
a) Under what condition is the second equation a consequence of the first?
b) Under what condition is the first equation a consequ... | Solution. a) Let $x_{0}$ be a root of the first equation, i.e., the numerical equality
$$
\frac{f_{1}\left(x_{0}\right)}{f_{2}\left(x_{0}\right)}=\frac{f_{3}\left(x_{0}\right)}{f_{4}\left(x_{0}\right)}
$$
holds. From this, we obtain the equality
$$
f_{1}\left(x_{0}\right) f_{4}\left(x_{0}\right)=f_{2}\left(x_{0}\rig... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,463 |
Example 16. Given two equations:
$$
f(x)=g(x) \quad \text { and } \quad \tan f(x)=\tan g(x)
$$
a) Can roots be lost as a result of the transition from the first equation to the second?
b) Can extraneous roots appear during this transition? | Solution. We will show that when transitioning from the first equation to the second, both the loss of roots and the acquisition of extraneous roots are possible.
a) Let \( f(x) = \arcsin x \) and \( g(x) = 2 \arcsin \frac{x}{\sqrt{2}} \). Since the number 1 is a root of the equation
\[
\arcsin x = 2 \arcsin \frac{x}... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,464 |
Example 17. Are the equations
$$
\operatorname{tg} 2 x-\operatorname{ctg} x=0 \quad \text { and } \quad \frac{2 \operatorname{tg} x}{1-\operatorname{tg}^{2} x}-\frac{1}{\operatorname{tg} x}=0
$$
equivalent? | Solution. The set of all solutions of the first equation consists of three series of solutions
\[
\begin{array}{ll}
x_{k}=\frac{\pi}{2}+\pi k, & k \in \mathbf{Z} \\
x_{m}=\frac{\pi}{6}+\pi m, & m \in \mathbf{Z} \\
x_{n}=\frac{-\pi}{6}+\pi n, & n \in \mathbf{Z}
\end{array}
\]
of which the series of solutions $x_{k}=\p... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,465 |
Example 18. Solve the equation
$$
\log _{2} x^{2}=2
$$ | Solution.
The first method. The domain of the equation is any $x \neq 0$. Applying the formula $\log _{2} x^{2}=2 \log _{2} x(x>0)$, we get the equation $2 \log _{2} x=2$, from which we find $x=2$.
The transformation performed allowed solving the equation (6) only on a part of its domain, specifically for positive $x... | x_{1}=2x_{2}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,466 |
Example 19. Solve the equation
$$
\sqrt{x+11}=x-1
$$ | Solution.
The first method. The domain of definition (DOD) of equation (7) is given by the condition $x \geqslant-11$. Considering that
$$
f(x)=g(x) \Rightarrow f^{2}(x)=g^{2}(x)
$$
we have
$$
\text { (7) } \begin{aligned}
\Rightarrow x+11=x^{2}-2 x+1 \Leftrightarrow x^{2}-3 x-10=0 \Leftrightarrow \\
\Leftrightarro... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,467 |
Example 20. Solve the equation
$$
\sqrt{x^{2}(x-1)}=|x|
$$ | Solution.
The first method. The domain of definition (DOD) of equation (8) is given by the condition $x^{2}(x-1) \geqslant 0$, i.e., $x=0$ and $x \geqslant 1$. Let's divide the DOD into two parts: $x=0$ and $x \geqslant 1$.
If $x=0$, then equation (8) turns into a true numerical equality. Thus, the number 0 is a solu... | x_{1}=0,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,468 |
Example 21. Solve the equation
$$
\sqrt{x-2}\left(x^{2}-4 x+3\right)=0
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,469 | ||
Example 2. Are the inequalities
$$
x+3-\frac{1}{x-1}>-x+2-\frac{1}{x-1} \text { and } x+3>-x+2
$$
equivalent? | Solution. The second inequality is obtained from the first inequality by adding the same expression $\frac{1}{x-1}$ to both of its parts, which is not defined at $x=1$. This means that the number $x=1$ cannot be a solution to the first inequality. However, $x=1$ is a solution to the second inequality. Therefore, there ... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,470 |
Example 4. Are the inequalities
$$
\sqrt{x-1}<\sqrt{2-x} \text { and } x-1<2-x
$$
equivalent? | Solution. The domain of the first inequality is defined by the system
$$
\left\{\begin{array}{l}
x-1 \geqslant 0 \\
2-x \geqslant 0
\end{array}\right.
$$
and, therefore, consists of all numbers in the interval $[1 ; 2]$.
The solution to the second inequality is all numbers from the interval $(-\infty ; 3 / 2)$.
Thu... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,472 |
Example 5. Are the inequalities
$$
\lg \left(x^{2}-4\right)>\lg (4 x-7) \quad \text { and } \quad x^{2}-4>4 x-7 \text { equivalent? }
$$ | Solution. Since
$$
x^{2}-4>4 x-7 \Leftrightarrow(x-1)(x-3)>0
$$
the set of solutions to the second of the given inequalities consists of all numbers in the intervals $(-\infty ; 1)$ and $(3 ;+\infty)$. However, for example, the number $x=0$ from the interval ( $-\infty ; 1$ ) is not a solution to the inequality $\lg ... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 40,473 |
Example 6. Prove that
$$
\left\{\begin{array}{l}
f(x)>0, \\
g(x)>0
\end{array} \Rightarrow \left\{\begin{array}{l}
f(x)>0 \\
f(x)+g(x)>0
\end{array}\right.\right.
$$
Provide an example when these systems are not equivalent. | Solution. Let the number $x_{0}$ be a solution to the first system; then the numerical inequalities $f\left(x_{0}\right)>0$ and $g\left(x_{0}\right)>0$ hold, and thus $f\left(x_{0}\right)+g\left(x_{0}\right)>0$. Therefore, the second system is a consequence of the first system.
Let $f(x)=x^{2}$ and $g(x)=x$. We will p... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 40,474 |
Example 7. Prove that
$$
\sqrt{f^{2}(x)} \geqslant \sqrt{g^{2}(x)} \Leftrightarrow|f(x)| \geqslant|g(x)| .
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 40,475 | |
Example 1. Solve the equation
$$
x^{2}-5|x|+6=0
$$ | Solution. The original equation is equivalent to the following systems:
$$
\left\{\begin{array} { l }
{ x ^ { 2 } - 5 x + 6 = 0 , } \\
{ x \geqslant 0 }
\end{array} \quad \left\{\begin{array}{l}
x^{2}+5 x+6=0 \\
x<0
\end{array}\right.\right.
$$
The equation $x^{2}-5 x+6=0$ has two solutions: $x_{1}=2, x_{2}=3$, each... | -2,2,-3,3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,476 |
Example 3. Find all roots of the equation
$$
2\left|x^{2}+2 x-5\right|=x-1
$$
satisfying the inequality $x<\sqrt{2}$. | Solution. The given equation is equivalent to the following system of equations
which on the set $(-\infty ; \sqrt{2})$ is equivalent to the following system
$$
\left\{\begin{array} { l }... | \frac{\sqrt{113}-5}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,478 |
Example 4. Solve the equation
$$
\left|\frac{x^{2}-6 \sqrt{x}+7}{x^{2}+6 \sqrt{x}+7}\right|=1
$$ | Solution. The given equation is equivalent to the system of equations
$\left[\begin{array}{l}\frac{x^{2}-6 \sqrt{x}+7}{x^{2}+6 \sqrt{x}+7}=1, \\ \frac{x^{2}-6 \sqrt{x}+7}{x^{2}+6 \sqrt{x}+7}=-1\end{array} \Leftrightarrow\right.$
$$
\Leftrightarrow\left[\begin{array}{l}
\frac{-12 \sqrt{x}}{x^{2}+6 \sqrt{x}+7}=0 \\
\fr... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,479 |
Example 5. Solve the equation
$$
\left|\frac{x^{2}-10 x+21}{x^{2}-12 x+32}\right|=-\frac{x^{2}-10 x+21}{x^{2}-12 x+32}
$$ | Solution. The given equation has the form $|f(x)|=-f(x)$, where $f(x)=\frac{x^{2}-10 x+21}{x^{2}-12 x+32}$. Such an equation is equivalent to the combination of systems
$$
\left\{\begin{array}{l}
f(x) = -f(x), \\
f(x) \geqslant 0,
\end{array} \quad \left\{\begin{array}{l}
f(x)=f(x) \\
f(x)<0
\end{array}\right.\right.
... | [3;4)\cup[7;8) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,480 |
Example 6. Solve the equation
$$
\frac{1-2 x}{3-|x-1|}=1
$$ | Solution. The given equation is equivalent to the combination of two systems:
Solving the equation $\frac{1-2 x}{4-x}=1$, we find $x_{1}=-3$ - its only root. However, it does not satisfy th... | -\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,481 |
Example 7. Solve the equation
$$
|x-| 4-x||-2 x=4
$$ | Solution. The given equation is equivalent to the combination of two systems:
$$
\left\{\begin{array} { l }
{ 4 - x \geqslant 0 } \\
{ | x - ( 4 - x ) | - 2 x = 4 , }
\end{array} \quad \left\{\begin{array}{l}
4-x4 \\
-2 x=0
\end{array}\right.\right.
$$
The second system of the combination (4) has no solutions.
The ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,482 |
Example 8. Solve the equation
$$
|3 x-8|-|3 x-2|=6
$$ | Solution. Using the interval method (Fig. 2.2), we find the intervals of sign constancy of the expressions $3 x-8$ and $3 x-2: x8 / 3$.
Thus, equation (7) is equivalent to the union of three systems:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x \leqslant 2 / 3 , } \\
{ - ( 3 x - 8 ) + ( 3 x - 2 ) = 6 , }
\end... | (-\infty;2/3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,483 |
Example 9. Solve the equation
$$
|x|+|7-x|+2|x-2|=4
$$ | Solution. Using the interval method, we find (Fig. 2.3) the intervals of sign constancy for the expressions \(x, 7-x\), and \(x-2: x7\).
Fig. 2.3
Thus,
}{x-1}=\frac{x}{x-1}+x \\
& \frac{x^{2}}{|x-1|}=\frac{\left|x^{2}\right|}{|x-1|}=\left|\frac{x^{2}}{x-1}\right|
\end{aligned}
$$
the equation (18) can be rewritten as
$$
\left|\frac{x}{x-1}\right|+|x|=\left|\frac{x}{x-1}+x\rig... | 0orx\in(1,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,486 |
Example 13. Solve the system of equations
$$
\left\{\begin{array}{l}
|x+y-4|=5 \\
|x-3|+|y-1|=5
\end{array}\right.
$$ | Solution. The given system is equivalent to the system
$$
\left\{\begin{array}{l}
|x+y-4|=5 \\
|x-3|+|y-1|=5 \\
|x+y-4|=|x-3|+|y-1|
\end{array}\right.
$$
which (since $x+y-4=(x-3)+(y-1)$) by statement 2 is equivalent to the mixed system
$$
\left\{\begin{array}{l}
|x+y-4|=5 \\
|x-3|+|y-1|=5 \\
(x-3)(y-1) \geqslant 0
... | \begin{aligned}&(;9-),\quad\text{where}\in[3;8];\\&(-1-),\text{where}\in[-2;3]0\end{aligned} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,487 |
Example 1. Prove that the equation has no solutions:
a) $\sqrt{x+2}=-2 ; \quad$ b) $\sqrt{2 x+3}+\sqrt{x+3}=0$;
c) $\sqrt{4-x}-\sqrt{x-6}=2$; d) $\sqrt{-1-x}=\sqrt[3]{x-5}$;
e) $5 \sqrt{x}-3 \sqrt{-x}+\frac{17}{x}=4 ; \quad$ f) $\sqrt{x-3}-\sqrt{x+9}=\sqrt{x-2}$;
g) $\sqrt{x}+\sqrt{x+9}=2 ; \quad$ h) $\sqrt[3]{x+1 ... | Solution. a) The arithmetic root cannot be a negative number, therefore the equation has no solutions.
b) The left side of the original equation is defined for $x \geqslant -3 / 2$. For each such $x$, the value $\sqrt{2 x+3}$ is non-negative, and the value $\sqrt{x+3}$ is positive. Therefore, their sum is always great... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,488 |
Example 2. Solve the equation:
a) $\sqrt{1+3 x}=1-x$
b) $\sqrt{1+3 x}=x-1$ | Solution. a) Squaring both sides of the equation, we get $1+3 x=1-2 x+x^{2}$, i.e., the equation
$$
x^{2}-5 x=0
$$
which is a consequence of the original equation. We find the roots of this equation: $x_{1}=0$ and $x_{2}=5$.
Let's check if these roots satisfy the original equation.
Let $x=0$; then the original equa... | x_{1}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,489 |
Example 3. Solve the equation
$$
\sqrt{1+3 x}=x+1
$$ | Solution. Squaring both sides of the equation, we get
$$
1+3 x=x^{2}+2 x+1
$$
i.e., the equation
$$
x^{2}-x=0
$$
which is a consequence of the original equation.
Let's find its roots: $x_{1}=0$ and $x_{2}=1$. Substituting each of the found roots into the original equation, we verify that both are roots of the orig... | x_{1}=0,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,490 |
Example 4. Solve the equation
$$
3 \sqrt{x+3}-\sqrt{x-2}=7
$$ | Solution. Isolate one of the roots on the left side:
$$
3 \sqrt{x+3}=\sqrt{x-2}+7
$$
By squaring both sides of the obtained equation, we have
$$
9(x+3)=x-2+14 \sqrt{x-2}+49
$$
Combining like terms and isolating the radical on the right side, we get the equation
$$
4 x-10=7 \sqrt{x-2}
$$
By squaring both sides of ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,491 |
Example 5. Solve the equation
$$
\sqrt{11 x+3}-\sqrt{2-x}-\sqrt{9 x+7}+\sqrt{x-2}=0
$$ | Solution. Isolate two radicals on each side so that after squaring we get the simplest equation:
$$
\sqrt{11 x+3}-\sqrt{2-x}=\sqrt{9 x+7}-\sqrt{x-2}
$$
Perform a chain of transformations:
$$
\begin{gathered}
11 x+3-2 \sqrt{(11 x+3)(2-x)}+2-x= \\
=9 x+7-2 \sqrt{(9 x+7)(x-2)}+x-2 \\
\sqrt{6+19 x-11 x^{2}}=\sqrt{9 x^{2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,492 |
Example 6. Solve the equation
$$
\sqrt[3]{2 x-1}+\sqrt[3]{x-1}=1
$$ | Solution. Raising both sides of the equation to the third power, we get
$$
3 x-2+3 \sqrt[3]{(2 x-1)(x-1)}(\sqrt[3]{2 x-1}+\sqrt[3]{x-1})=1
$$
By the condition, the expression $\sqrt[3]{2 x-1}+\sqrt[3]{x-1}$ equals one. Substituting this expression with one in the obtained equation, we get the equation
$$
3 x-2+3 \sq... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,493 |
Example 7. Solve the equation
$$
\sqrt{2 x^{2}+5 x-2}-\sqrt{2 x^{2}+5 x-9}=1
$$ | Solution. First method. By direct verification, we find that the following relation holds:
$$
\left(\sqrt{2 x^{2}+5 x-2}\right)^{2}-\left(\sqrt{2 x^{2}+5 x-9}\right)^{2}=7
$$
We factorize the left-hand side of this equation:
$$
\begin{aligned}
&\left(\sqrt{2 x^{2}+5 x-2}-\sqrt{2 x^{2}+5 x-9}\right) \times \\
& \time... | x_{1}=2,x_{2}=-9/2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,494 |
Example 8. Solve the equation
$$
\frac{1}{x}+\frac{1}{\sqrt{1-x^{2}}}=\frac{35}{12}
$$ | Solution. By squaring both sides of the equation and combining like terms, we obtain the equation
$$
\frac{1}{x^{2}\left(1-x^{2}\right)}+\frac{2}{x \sqrt{1-x^{2}}}-\left(\frac{35}{12}\right)^{2}=0
$$
which is a consequence of the original equation.
$$
\begin{aligned}
\text { Let } y=\frac{1}{x \sqrt{1-x^{2}}} ; \tex... | {-\frac{5+\sqrt{73}}{14},\frac{3}{5},\frac{4}{5}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,495 |
Example 9. Solve the equation
$$
\sqrt{(x-1)^{2}(x-4)}=|x-1| \sqrt{16-x^{2}} .
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{l}
(x-1)^{2}(x-4) \geqslant 0 \\
16-x^{2} \geqslant 0
\end{array}\right.
$$
we find the domain of the equation: $x_{1}=1, x_{2}=4$. By verification, we establish that each of them is a root of equation (3).
Thus, the set of all roots of equation (3... | x_{1}=1x_{2}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,496 |
Example 10. Solve the equation
$$
\sqrt{x+16}-x+4=0
$$ | Solution. The domain of the equation: $x \geqslant-16$. Isolating the radical, we get the equation $\sqrt{x+16}=x-4$. Squaring both sides of this equation and combining like terms, we obtain the equation $x^{2}-9 x=0$, the roots of which are: $x_{1}=0$ and $x_{2}=9$. Each of these roots belongs to the domain of the ori... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,497 |
Example 11. Solve the equation
$$
\sqrt{x^{2}+3 x-4}=\sqrt{2 x+2}
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}+3 x-4 \geqslant 0 \\
2 x+2 \geqslant 0
\end{array}\right.
$$
we find the domain of the equation: $x \geqslant 1$.
By squaring both sides of the original equation, we obtain the equation
$$
x^{2}+3 x-4=2 x+2
$$
which is a consequence of i... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,498 |
Example 12. Solve the equation
$$
\sqrt{2 x+5}+\sqrt{x-1}=8
$$ | Solution. Domain of the equation: $x \geqslant 1$. Squaring both sides of the equation and performing transformations, we obtain the equation
$$
2 \sqrt{2 x^{2}+3 x-5}=60-3 x
$$
After squaring both sides of this equation, we get the equation
$$
4\left(2 x^{2}+3 x-5\right)=(60-3 x)^{2}
$$
which is a consequence of t... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,499 |
Example 13. Solve the equation
$$
2 \sqrt{1-x^{2}}=x-2
$$ | Solution. Solving the inequality $1-x^{2} \geqslant 0$, we find the domain of the equation: $-1 \leqslant x \leqslant 1$.
By squaring both sides of the equation and combining like terms, we obtain the equation
$$
5 x^{2}-4 x=0
$$
which is a consequence of the original equation. The roots of this equation are: $x_{1}... | nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,500 |
Example 14. Solve the equation
$$
\sqrt{4 x+9}-\sqrt{11 x+1}-\sqrt{7 x+4}=0
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{r}
4 x+9 \geqslant 0 \\
11 x+1 \geqslant 0 \\
7 x+4 \geqslant 0
\end{array}\right.
$$
we find the domain of the equation: $x \geqslant-1 / 11$.
In this equation, any of the three radicals can be isolated. It is reasonable to isolate the radical $\s... | x_{1}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,501 |
Example 15. Solve the equation
$$
\sqrt{x+7} \sqrt{3 x-2}=3 \sqrt{x-1} \sqrt{x+2}
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{r}
x+7 \geqslant 0 \\
3 x-2 \geqslant 0 \\
x-1 \geqslant 0 \\
x+2 \geqslant 0
\end{array}\right.
$$
we find the domain of the equation: $x \geqslant 1$. Squaring both sides of the equation and combining like terms, we obtain the equation
$$
3 x^{2}... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,502 |
Example 16. Solve the equation
$$
\frac{3 x-2}{\sqrt{2 x-1}}=\sqrt{(2 x-1)^{3}}
$$ | Solution. Solving the inequality $2 x-1>0$, we find the domain of the equation: $x>1 / 2$.
Multiplying both sides of the original equation by $\sqrt{2 x-1}$ and simplifying it, we obtain the equation
$$
4 x^{2}-7 x+3=0
$$
which is a consequence of the original equation. Its roots are: $x_{1}=1$ and $x_{2}=3 / 4$.
B... | x_{1}=1,x_{2}=3/4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,503 |
Example 17. Solve the equation
$$
\frac{2}{2-\sqrt{x}}+\frac{1}{2}=\frac{4}{2 \sqrt{x}-x}
$$ | Solution. The domain of the equation is determined by the system of inequalities
$$
\left\{\begin{array}{r}
x \geqslant 0 \\
2-\sqrt{x} \neq 0 \\
2 \sqrt{-} x-x \neq 0
\end{array}\right.
$$
i.e., $x>0, x \neq 4$.
Multiplying both sides of the original equation by the expression $2(2 \sqrt{ } \bar{x}-x)$, we obtain t... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,504 |
Example 18. Solve the equation
$$
x=(\sqrt{1+x}+1)(\sqrt{10+x}-4)
$$ | Solution. The domain of definition (DOD) of equation (6): $x \geqslant-1$. Multiplying both sides of equation (6) by the expression $\sqrt{1+x}-1$, we obtain the equation
$$
x(\sqrt{1+x}-\sqrt{10+x}+3)=0
$$
which is a consequence of equation (6). This equation has two roots: $x_{1}=0$ and $x_{2}=-1$ (note that the DO... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,505 |
Example 19. Solve the equation
$$
\sqrt[3]{x^{2}-2}=\sqrt{2-x^{3}}
$$ | Solution. Solving the inequality $2-x^{3} \geqslant 0$, we find the domain of the original equation: $x \leqslant \sqrt[3]{2}$.
The right-hand side of the original equation is a non-negative number for any $x$ from the domain, so $x^{2}-2 \geqslant 0$, i.e., $|x| \geqslant \sqrt{2}$.
From the inequalities $x \leqslan... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,506 |
Example 21. Solve the equation
$$
\sqrt{x^{2}-1}=(x+5) \sqrt{\frac{x+1}{x-1}}
$$ | Solution. Using formulas 4 and 5, we obtain the equation
$$
\sqrt{|x+1|}\left(\sqrt{|x-1|}-(x+5) \frac{1}{\sqrt{|x-1|}}\right)=0
$$
which is a consequence of the original equation. Solving this equation reduces to solving the system of equations
$$
\sqrt{|x+1|}=0, \quad \sqrt{|x-1|}-\frac{x+5}{\sqrt{|x-1|}}=0
$$
Fr... | x_{1}=-1,x_{2}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,508 |
Example 22. Solve the equation
$$
\frac{\sqrt{2-x}}{\sqrt{2+x}}=\frac{2-x}{2+x}
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{l}
2-x \geqslant 0 \\
x+2>0
\end{array}\right.
$$
we find the domain of the given equation - the interval $-2<x \leqslant 2$.
Letting $t=\frac{\sqrt{2-x}}{\sqrt{2+x}}$, we get $t^{2}=\frac{2-x}{2+x}$; therefore, the given equation can be written as... | x_{1}=-2x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,509 |
Example 23. Solve the equation
$$
(x-3)(x+1)+3(x-3) \sqrt{\frac{x+1}{x-3}}-28=0
$$ | Solution. Solving the inequality $\frac{x+1}{x-3} \geqslant 0$, we find the domain of the equation (9): $x \leqslant-1$ and $x>3$.
Within the domain of the equation, we have:
$$
3(x-3) \sqrt{\frac{x+1}{x-3}}=3 \sqrt{(x+1)(x-3)} \text { for } x>3
$$
And
$$
3(x-3) \sqrt{\frac{x+1}{x-3}}=-3 \sqrt{(x+1)(x-3)} \text { f... | 1-\sqrt{53},1+2\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,510 |
Example 24. Solve the equation
$$
\sqrt[3]{x+1}-\sqrt[3]{x-1}=\sqrt[6]{x^{2}-1}
$$ | Solution. Solving the inequality $x^{2}-1 \geqslant 0$, we find the domain of the equation (10) - intervals $-\infty < x \leq -1$ and $x \geq 1$; in this case:
if $x < -1$ or $x > 1$, then $\sqrt[3]{x+1}=\sqrt[6]{(x+1)^{2}}$ and $\sqrt[3]{x-1}=\sqrt[6]{(x-1)^{2}}$
Therefore, on this part of the domain, dividing both ... | {-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,511 |
Example 25. Solve the equation
$$
\sqrt{4-6 x-x^{2}}=x+4
$$ | Solution. The given equation is equivalent to the system
$$
\left\{\begin{array}{l}
x+4 \geqslant 0 \\
4-6 x-x^{2}=(x+4)^{2}
\end{array}\right.
$$
i.e., the system
$$
\left\{\begin{array}{l}
x \geqslant-4 \\
x^{2}+7 x+6=0
\end{array}\right.
$$
## 68
Solving the equation \(x^{2}+7 x+6=0\), we find its two roots: \(... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,512 |
Example 26. Solve the equation
$$
\sqrt{x+5}+\sqrt{2 x+8}=7
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{r}
x+5 \geqslant 0 \\
2 x+8 \geqslant 0
\end{array}\right.
$$
we find the domain of the equation (12): $x \geqslant-4$. We have:
$$
\text { (12) } \begin{aligned}
\Leftrightarrow & \left\{\begin{array}{l}
x \geqslant-4 \\
x+5+2 \sqrt{x+5} \sqrt{2 x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,513 |
Example 27. Solve the equation
$$
\sqrt{5 x+7}-\sqrt{2 x+3}=\sqrt{3 x+4}
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{l}
5 x+7 \geqslant 0 \\
2 x+3 \geqslant 0 \\
3 x+4 \geqslant 0
\end{array}\right.
$$
we find the domain of the equation: $x \geqslant -4 / 3$. We have:
$$
\begin{aligned}
& (13) \Leftrightarrow\left\{\begin{array}{l}
x \geqslant -4 / 3 \\
\sqrt{5 x... | x_{1}=-\frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,514 |
Example 28. Solve the equation
$$
\sqrt{2 x^{2}-1}+\sqrt{x^{2}-3 x-2}=\sqrt{2 x^{2}+2 x+3}+\sqrt{x^{2}-x+2}
$$ | Solution. Rewrite equation (14) in such a way that after squaring both sides, we obtain an equivalent equation that is simpler than in other cases:
$$
\sqrt{2 x^{2}-1}-\sqrt{x^{2}-x+2}=\sqrt{2 x^{2}+2 x+3}-\sqrt{x^{2}-3 x-2}
$$
Considering that the left and right parts of the last equation must be simultaneously eith... | x_{1}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,515 |
Example 29. Solve the equation
$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$ | Let's denote $\sqrt{x-1}=t$, then we have $t \geqslant 0$ and $x=t^{2}+1$. By substituting the unknown in equation (17), we get
$$
\begin{aligned}
& \left\{\begin{array}{l}
t \geqslant 0 \\
\sqrt{t^{2}-4 t+4}+\sqrt{t^{2}-6 t+9}=1
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l }
{ t \g... | 5\leqslantx\leqslant10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,516 |
Example 30. Solve the equation
$$
\sqrt[3]{(x+1)^{2}}+2 \sqrt[3]{(x-1)^{2}}=3 \sqrt[3]{x^{2}-1}
$$ | Solution. The domain of definition (DOD) of equation (18) is the set of all real numbers. Note that if we let $u=\sqrt[3]{x+1}, v=$ $=\sqrt[3]{x-1}$, then equation (18) will take the form
$$
u^{2}+2 v^{2}-3 u v=0
$$
The left side of this equation is a homogeneous function of degree 2.
The number $x=1$ is not a root ... | \frac{9}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,517 |
Example 31. Solve the equation
$$
\sqrt[3]{8+x}+\sqrt[3]{8-x}=1
$$ | Solution. The domain of the equation is the set of all real numbers. Let $u=\sqrt[3]{8+x}, v=\sqrt[3]{8-x}$. Since $u^{3}+v^{3}=16$, for $u$ and $v$ we obtain the symmetric system of equations
$$
\left\{\begin{array}{c}
u+v=1 \\
u^{3}+v^{3}=16
\end{array}\right.
$$
solving which we find:
$$
\left\{\begin{array} { l ... | \3\sqrt{21} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,518 |
Example 32. Solve the equation
$$
\sqrt{1-x^{2}}=\left(\frac{2}{3}-\sqrt{\bar{x}}\right)^{4}
$$ | Solution. Let $\sqrt{x}=u \quad$ and $\quad 2 / 3-\sqrt{x}=v, \quad$ then we have $1-x^{2}=1-u^{4}$ and $(2 / 3-\sqrt{\bar{x}})^{4}=v^{4}$. For $u$ and $v$, we get:
$$
\left\{\begin{array} { r l }
{ u ^ { 4 } + v ^ { 4 } } & { = 1 } \\
{ u + v } & { = 2 / 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2(u v)... | x_{1}=\frac{1}{9}(1+\sqrt{\sqrt{\frac{97}{2}}-3})^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,519 |
Example 33. Solve the system
$$
\left\{\begin{array}{l}
\sqrt{5 x(x+2)}+\sqrt{5 y(x+2)}=2 \sqrt{2(x+y)(x+2)} \\
x y=9
\end{array}\right.
$$ | The solution is as follows. We replace system (24) with the system
$$
\begin{cases}
\sqrt{5|x|} \sqrt{|x+2|} + \sqrt{5|y|} \sqrt{|x+2|} = 2 \sqrt{2|x+y|} \sqrt{|x+2|} \\
xy = 9
\end{cases}
$$
which is a consequence of system (24). System (25) is equivalent to the combination of two systems:
$$
\left\{\begin{array}{l... | (-9,-1);(-2,-9/2);(1,9);(9,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,520 |
Example 1. Solve the equation
$$
4^{x}=8^{2 x-3}
$$ | Solution. Since
$$
4^{x}=\left(2^{2}\right)^{x}=2^{2 x}, \quad 8^{2 x-3}=\left(2^{3}\right)^{2 x-3}=2^{6 x-9}
$$
then
$$
2^{2 x}=2^{6 x-9} \Leftrightarrow 2 x=6 x-9 \Leftrightarrow x=9 / 4
$$ | \frac{9}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,522 |
Example 2. Solve the equation
$$
(0.4)^{x-1}=(6.25)^{6x-5}
$$ | Solution. Since
$$
\begin{gathered}
(0.4)^{x-1}=(2 / 5)^{x-1} \\
(6.25)^{6 x-5}=(25 / 4)^{6 x-5}=(5 / 2)^{12 x-10}=(2 / 5)^{10-12 x}
\end{gathered}
$$
then
$$
(2 / 5)^{x-1}=(2 / 5)^{10-12 x} \Leftrightarrow x-1=10-12 x \Leftrightarrow x=11 / 13
$$
When solving simple exponential equations, a transformation is used ... | \frac{11}{13} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,523 |
Example 3. Solve the equation
$$
5^{2 x+1}-3 \cdot 5^{2 x-1}=550
$$ | Solution. Factoring out the expression $5^{2 x-1}$ on the left side of the equation, we get
$$
5^{2 x-1}\left(5^{2}-3\right)=550 \Leftrightarrow 5^{2 x-1}=5^{2} \Leftrightarrow 2 x-1=2 \Leftrightarrow x=3 / 2
$$ | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,524 |
Example 4. Solve the equation
$$
\sqrt{3^{x}} 5^{x / 2}=225
$$
The solution. Since $\sqrt{3^{x}} 5^{x / 2}=3^{x / 2} 5^{x / 2}=15^{x / 2}$, the original equation is equivalent to the equation $15^{x / 2}=15^{2}$, i.e., $x=4$.
An equation of the form
$$
a^{f(x)}=1, \quad a>0, \quad a \neq 1
$$
is equivalent to the ... | Solution. The given equation is equivalent to the equation $x^{2}+$ $+x-2=0$. Therefore, the given equation has two roots: $x_{1}=-2, x_{2}=1$. | x_{1}=-2,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,525 |
Example 6. Solve the equation
$$
\sqrt{5-x}\left(3^{x^{2}-7.2 x+3.9}-9 \sqrt{3}\right)=0
$$ | Solution. The domain of permissible values of equation (1) is determined by the condition $5-x \geqslant 0$, i.e., $x \leqslant 5$. For such values of $x$, equation (1) is equivalent to the system of equations
$$
\sqrt{5-x}=0, \quad 3^{x^{2}-7.2 x+3.9}=9 \sqrt{3}
$$
From the first equation, we find $x_{1}=5$.
To sol... | 5,\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,526 |
Example 8. Solve the equation
$$
4^{x}-3^{x-1 / 2}=3^{x+1 / 2}-2^{2 x-1}
$$ | Solution. Group the terms containing powers with base 4 and base 3:
$$
4^{x}+(1 / 2) 4^{x}=3^{x+1 / 2}+3^{x-1 / 2}
$$
Factor out the common factors:
$$
4^{x}(1+1 / 2)=3^{x-1 / 2}(1+3)
$$
Divide this equation by the expression on its right side, we get $(4 / 3)^{x-3 / 2}=1$. From this, we find $x-3 / 2=0$; hence, $x... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,527 |
Example 9. Solve the equation
$$
5^{2 x}-2 \cdot 5^{x}-15=0
$$ | Solution. Let $t=5^{x}$. Then $t^{2}-2 t-15=0$. From this, we find $t_{1}=5, t_{2}=-3$. Thus, the given equation is equivalent to the system of equations
$$
5^{x}=5, \quad 5^{x}=-3
$$
## § 3. EXPONENTIAL EQUATIONS
The second equation of this system has no roots, since $-3<0$ for any $x$, while from the first equatio... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,528 |
Example 10. Solve the equation
$$
5^{2 x-1}=7^{3-x}
$$ | Solution. Both sides of the equation are positive; therefore, it can be logarithmized with base 5. We obtain the equation
$$
2 x-1=(3-x) \log _{5} 7
$$
equivalent to the original one.
Thus,
$$
x\left(2+\log _{5} 7\right)=1+3 \log _{5} 7
$$
i.e., $x=\frac{1+3 \log _{5} 7}{2+\log _{5} 7}$. | \frac{1+3\log_{5}7}{2+\log_{5}7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,529 |
Example 11. Solve the equation
$$
\sqrt{\frac{3}{5}}\left(\frac{3}{5}\right)^{x-1}=\frac{\sqrt[4]{5^{3 x-4}}}{\sqrt{5}}
$$ | Solution. Both sides of the given equation are positive. By taking the logarithm of both sides of the equation to the base 5, we obtain the equation
$$
(x-1) \log _{5}\left(\frac{3}{5}\right)+\frac{1}{2} \log _{5}\left(\frac{3}{5}\right)=\frac{3 x-4}{4}-\frac{1}{2}
$$
i.e., the equation
$$
x\left(\log _{5} 3-1\right... | \frac{2(3\log_{5}3-4)}{4\log_{5}3-7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,530 |
Example 12. Solve the equation
$$
8^{2 / x}-2^{(3 x+3) / x}+12=0
$$ | The solution is as follows. Since
$$
2^{(3 x+3) / x}=2^{3+3 / x}=8 \cdot 2^{3 / x}=8 \cdot\left(2^{3}\right)^{1 / x}=8 \cdot 8^{1 / x},
$$
the equation (5) is equivalent to the equation
$$
8^{2 / x}-8 \cdot 8^{1 / x}+12=0
$$
Let $8^{1 / x}=t$; then the equation (5) is equivalent to the system of equations
$$
8^{1 ... | x_{1}=3\log_{6}2,x_{2}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,531 |
Example 13. Solve the equation
$$
3 \cdot 16^{x}+37 \cdot 36^{x}=26 \cdot 81^{x}
$$ | Solution. In this equation, the numbers $16,36,81$ form three consecutive terms of a geometric progression (with a common ratio of $9 / 4$).
To solve the original equation, divide both sides by $81^{x}$. We get
$$
3 \cdot(4 / 9)^{2 x}+37 \cdot(4 / 9)^{x}-26=0
$$
Let $t=(4 / 9)^{x}$; then equation (8) becomes
$$
3 t... | x_{1}=1/2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,532 |
Example 14. Solve the equation
$$
3^{2 x^{2}+6 x-9}+4 \cdot 15^{x^{2}+3 x-5}=3 \cdot 5^{2 x^{2}+6 x-9}
$$ | Solution. Using the properties of exponents, we can represent the given equation as
$$
\frac{1}{3} \cdot 9^{x^{2}+3 x-4}+\frac{4}{15} \cdot 15^{x^{2}+3 x-4}=\frac{3}{5} 25^{x^{2}+3 x-4}
$$
Divide both sides of equation (10) by $15^{x^{2}+3 x-4}$ and introduce the notation $y=(3 / 5)^{x^{2}-3 x-4}$. We obtain the equa... | x=-4x=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,533 |
Example 15. Solve the equation
$$
27^{x}+12^{x}=2 \cdot 8^{x}
$$ | Solution. This equation is similar in appearance to equation (6): the exponent of the bases is the same, but the bases 27, 12, and 8 of the three consecutive terms of a geometric progression do not form one.
Consecutive (but four) terms of a geometric progression are the numbers $27, 18, 12$, and 8. Therefore, we can ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,534 |
Example 16. Solve the equation
$$
5 \cdot 2^{3 x-3}-3 \cdot 2^{5-3 x}+7=0
$$ | Solution. Using the properties of exponents, rewrite the given equation as
$$
\frac{5}{8} \cdot 2^{3 x}-\frac{96}{2^{3 x}}+7=0
$$
This equation is of the form (12).
Let $t=2^{3 x}$; then we have $\frac{5}{8} t-\frac{96}{t}+7=0$, i.e., $5 t^{2}+56 t-768=0$. From this, we find $t_{1}=-96 / 5, t_{2}=8$.
Thus, equation... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,535 |
Example 17. Solve the equation
$$
(2+\sqrt{3})^{x^{2}-2 x+1}+(2-\sqrt{3})^{x^{2}-2 x-1}=\frac{101}{10(2-\sqrt{3})}
$$ | Solution. Since $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$, multiplying both sides of the equation by $2-\sqrt{3}$, we transform it into the form (12):
$$
(2+\sqrt{3})^{x^{2}-2 x}+\frac{1}{(2+\sqrt{3})^{x^{2}-2 x}}=\frac{101}{10}
$$
Let $t=(2+\sqrt{3})^{x^{2}-2 x}$; then equation (15) takes the form
$$
t+1 / t=101 / 10
$$
T... | x_{1}=1+\sqrt{1+\log_{2+\sqrt{3}}10},\quadx_{2}=1-\sqrt{1+\log_{2+\sqrt{3}}10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,536 |
Example 18. Solve the equation
$$
3^{2 x^{2}-6 x+3}+6^{x^{2}-3 x+1}=2^{2 x^{2}-6 x+3}
$$ | Solution. Equation (17) is equivalent to the equation
$$
27 \cdot 3^{2\left(x^{2}-3 x\right)}+6 \cdot 3^{x^{2}-3 x} 2^{x^{2}-3 x}-8 \cdot 2^{2}\left(x^{2}-3 x\right)=0
$$
the solution of which reduces to solving the homogeneous equation
$$
27 f^{2}(x)+6 f(x) g(x)-8 g^{2}(x)=0
$$
where $f(x)=3^{x^{2}-3 x}, g(x)=2^{x... | x_{1}=1,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,537 |
Example 19. Solve the equation
$$
6 \sqrt[x]{9}-13 \sqrt[x]{6}+6 \sqrt[x]{4}=0
$$ | Solution. The domain of admissible values of the given equation consists of all natural numbers greater than 1.
Equation (19) is homogeneous; dividing both sides of the equation by $\sqrt[x]{\overline{4}}$ and setting $t=\sqrt[x]{3 / 2}$, we obtain the equation
$$
6 t^{2}-13 t+6=0
$$
the roots of which are $t_{1}=3 ... | noroots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,538 |
Example 20. Solve the equation
$$
3^{x-1}+5^{x-1}=34
$$ | Solution. Note that $3^{2}+5^{2}=34$. Therefore, equation (20) has a solution $x_{1}=3$. We will prove that there are no other solutions.
Indeed, each of the functions $y=3^{x-1}$ and $y=5^{x-1}$, being exponential functions with a base greater than 1, is increasing; therefore, their sum is also an increasing function... | x_1=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,539 |
Example 21. Solve the equation
$$
1+3^{x / 2}=2^{x}
$$ | Solution. Dividing both sides of the equation by $2^{x}$, we obtain the equation
$$
(1 / 2)^{x}+(\sqrt{\overline{3}} / 2)^{x}=1
$$
equivalent to equation (21). Equation (22) can be written as
$$
\left(\sin \frac{\pi}{6}\right)^{x}+\left(\cos \frac{\pi}{6}\right)^{x}=1
$$
Comparing this equation with the fundamental... | x_1=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,540 |
Example 22. Solve the equation
$$
3^{x} \cdot 8^{x /(x+1)}=36
$$ | Solution. We have:
$$
\begin{aligned}
3^{x} \cdot 2^{3 x /(x+1)}=3^{2} \cdot 2^{2} \Leftrightarrow & 3^{x-2} \cdot 2^{3 x /(x+1)-2}=1 \Leftrightarrow \\
& \Leftrightarrow 3^{x-2} \cdot 2^{(x-2) /(x+1)}=1 \Leftrightarrow\left(3 \cdot 2^{1 /(x+1)}\right)^{x-2}=1
\end{aligned}
$$
Thus, equation (24) is equivalent to the... | x_{1}=2,x_{2}=-1-\lg_{3}2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,541 |
Example 23. Solve the equation
$$
x \sqrt[3]{x^{2}}=(\sqrt{x})^{x}
$$ | Solution. The domain of admissible values of the given equation is determined by the condition $x>0$. For positive $x$, equation (25) is equivalent to the equation
$$
10^{x^{2 / 3} \cdot \lg x}=10^{(x \lg x) / 2}
$$
i.e., the equation
$$
\left(x^{2 / 3}-x / 2\right) \lg x=0
$$
From this, it follows that for $x>0$, ... | x_{1}=8,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,542 |
Example 24. Solve the system
$$
\left\{\begin{array}{l}
(3 / 2)^{x-y}-(2 / 3)^{x-y}=65 / 36 \\
x y-x+y=118
\end{array}\right.
$$ | Solution. Denoting $(3 / 2)^{x-y}=t$, the first equation of the given system can be written as
$$
t-\frac{1}{t}=\frac{65}{36}
$$
This equation is equivalent to the equation
$$
\frac{36 t^{2}-65 t-36}{36 t}=0
$$
i.e., the equation
$$
\frac{36(t-9 / 4)(t+4 / 9)}{36 t}=0
$$
Since $t>0$, the last equation has a uniqu... | (12;10)(-10;-12) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,543 |
Example 25. Solve the system
$$
\left\{\begin{array}{l}
y^{x^{2}+7 x+12}=1 \\
x+y=6 \\
y>0
\end{array}\right.
$$ | Solution. The first equation under the condition $y>0$ is equivalent to the system of equations
$$
y=1, \quad x^{2}+7 x+12=0
$$
i.e., the system of equations
$$
y=1, \quad x=-3, \quad x=-4
$$
Therefore, the given system is equivalent to the combination of three systems:
$$
\left\{\begin{array} { l }
{ y = 1 , } \... | (5;1),(-3;9),(-4;10) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,544 |
Example 1. Solve the equation:
a) $1-\lg 5=\frac{1}{3}\left(\lg \frac{1}{2}+\lg x+\frac{1}{3} \lg 5\right)$;
b) $\log _{1 / 3}(-1 / x)=2 ; \quad$ c) $\lg (2 x-5)^{2}=0$. | Solution. a) Since $1-\lg 5=\lg 10-\lg 5=\lg 2, \lg \frac{1}{2}=$ $=-\lg 2$, the original equation is equivalent to the equation
$$
4 \lg 2-\frac{1}{3} \lg 5=\lg x
$$
i.e., the equation $\lg \frac{16}{\sqrt[3]{5}}=\lg x$. From this, we obtain $x=\frac{16}{\sqrt[3]{5}}$ - the only root of the given equation.
b) The o... | \frac{16}{\sqrt[3]{5}},-9,x_{1}=3,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,546 |
Example 2. Solve the equation:
a) $0.2 \log _{x} \frac{1}{32}=-0.5$;
b) $\log _{x-1} 3=2$;
c) $\log _{\log _{3}} x=2$. | Solution. a) Since $\frac{1}{32}=\frac{1}{2^{5}}=\left(\frac{1}{2}\right)^{5}$, then $\log _{x} \frac{1}{32}=$ $=5 \log _{x} \frac{1}{2}$, and therefore, the original equation is equivalent to the equation
$$
\log _{x} \frac{1}{2}=-\frac{1}{2}
$$
from which $x=(1 / 2)^{-2}=4$. The number 4 is the only root of the ori... | 4,1+\sqrt{3},3^{\sqrt{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,547 |
Example 3. Prove that the equation has no solutions:
a) $\log _{3}(3+\sqrt{ } \bar{x})+\log _{3}\left(1+x^{2}\right)=0$;
b) $\log _{1 / 3}(1+\sqrt{ } \bar{x})+\log _{1 / 3}(1+x)=2$;
c) $\lg (4-x)-\lg (x-6)=5$;
d) $\lg (x-3)-\lg (x+9)=\lg (x-2)$;
e) $\log _{1 / 10}(x-3)-\log _{1 / 10}(x+9)=\log _{1 / 10}(x-2)$;
f)... | Solution. a) The domain of definition (ODZ) of the equation is determined by the inequality $x \geqslant 0$. On the ODZ, we have
$$
3+\sqrt{x} \geqslant 3 \quad \text { and } \quad 1+x^{2} \geqslant 1 ;
$$
therefore, $\log _{3}(3+\sqrt{x}) \geqslant 1$ and $\log _{3}\left(1+x^{2}\right) \geqslant 0$. The sum of a pos... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,548 |
Example 5. Solve the equation:
a) $\log _{1 / 5} \frac{2+x}{10}=\log _{1 / 5} \frac{2}{x+1}$;
b) $\log _{3}\left(x^{2}-4 x+3\right)=\log _{3}(3 x+21)$;
c) $\log _{1 / 10} \frac{2 x^{2}-54}{x+3}=\log _{1 / 10}(x-4)$;
d) $\log _{(5+x) / 3} 3=\log _{-1 /(x+1)} 3$. | Solution. a) Equation (6) is equivalent to the system
$$
\left\{\begin{array}{l}
x+1>0 \\
\frac{2+x}{10}=\frac{2}{x+1}
\end{array}\right.
$$
The equation of the system has two roots: $x_{1}=3, x_{2}=-6$. However, only the number $x_{1}=3$ satisfies the first condition. Thus, equation (6) has a single root, the number... | 3,-2,9,6,-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,549 |
Example 6. Solve the equation:
a) $\log _{x+1}\left(x^{2}-3 x+1\right)=1$
b) $\log _{x}\left(2 x^{2}-3 x-4\right)=2$. | Solution. a) The given equation is equivalent to the system
$$
\left\{\begin{array} { l }
{ x + 1 > 0 } \\
{ x + 1 \neq 1 , } \\
{ x ^ { 2 } - 3 x + 1 = x + 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>-1 \\
x \neq 0 \\
x^{2}-4 x=0
\end{array} \Leftrightarrow x=4\right.\right.
$$
Therefore, the only roo... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,550 |
Example 7. Solve the equation:
a) $\log _{x^{2}-1}\left(x^{3}+6\right)=\log _{x^{2}-1}\left(4 x^{2}-x\right)$;
b) $\log _{x^{3}+x}\left(x^{2}-4\right)=\log _{4 x^{2}-6}\left(x^{2}-4\right)$. | Solution. a) The given equation is equivalent to the system
$$
\left\{\begin{array}{l}
x^{3}+6>0 \\
x^{2}-1>0 \\
x^{2}-1 \neq 1 \\
x^{3}+6=4 x^{2}-x
\end{array}\right.
$$
The equation of this system $x^{3}-4 x^{2}+x+6=0$ has three roots: $x_{1}=-1, x_{2}=2, x_{3}=3$. The number $x_{1}=-1$ does not satisfy the conditi... | x_2=2,x_3=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,551 |
Example 9. Solve the equation:
$$
\lg 2 x=2 \lg (4 x-15)
$$ | Solution. The equation is equivalent to the system
$$
\left\{\begin{array} { l }
{ 4 x - 1 5 > 0 } \\
{ \operatorname { l g } 2 x = \operatorname { l g } ( 4 x - 1 5 ) ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
4 x-15>0 \\
2 x=(4 x-15)^{2}
\end{array}\right.\right.
$$
Consider the equation of the ... | \frac{9}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,552 |
Example 10. Solve the equation
$$
\lg \left(8-10 x-12 x^{2}\right)=3 \lg (2 x-1)
$$ | Solution. The equation is equivalent to the equation
$$
\lg \left(8-10 x-12 x^{2}\right)=\lg (2 x-1)^{3}
$$
which is equivalent to the system
$$
\left\{\begin{array}{l}
2 x-1>0 \\
8-10 x-12 x^{2}=(2 x-1)^{3}
\end{array}\right.
$$
After transformations, the equation of the system takes the form
$$
(2 x-1)\left(4 x^... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,553 |
Example 11. Solve the equation
$$
\lg x=\frac{1}{2} \lg (x+1)
$$ | Solution. The equation is equivalent to the equation
$$
2 \lg x=\lg (x+1)
$$
which is equivalent to the system
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x > 0 , } \\
{ \operatorname { l g } x ^ { 2 } = \operatorname { l g } ( x + 1 ) }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ x > 0 , } \\
{... | \frac{\sqrt{5}+1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,554 |
Example 12. Solve the equation:
a) $\log _{3}(x-2)+\log _{3} x=\log _{3} 8$;
b) $\lg (x-9)+2 \lg \sqrt{2 x-1}=2$;
B) $\frac{1}{2} \log _{5}(x+5)+\log _{5} \sqrt{x-3}=\frac{1}{2} \log _{5}(2 x+1)$. | Solution. a) The equation is equivalent to the system
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x - 2 > 0 , } \\
{ x > 0 , } \\
{ \operatorname { log } _ { 3 } ( x ( x - 2 ) ) = \operatorname { log } _ { 3 } 8 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>2, \\
x>0, \\
x(x-2)=8
\end{array} \Leftrigh... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,555 |
Example 14. Solve the equation:
a) $2 \log _{2} x+\log _{\sqrt{2}} x+\log _{1 / 2} x=9$;
b) $\log _{\sqrt{2}}^{2} x+3 \log _{2} x+\log _{1 / 2} x=2$;
c) $\log _{5}^{2} x+\log _{5 x} \frac{5}{x}=1$ | Solution. a) The domain of the equation is the interval $x>0$. Since
$$
\begin{aligned}
& \log _{\sqrt{2}} x=\frac{\log _{2} x}{\log _{2} \sqrt{2}}=2 \log _{2} x \\
& \log _{1 / 2} x=\frac{\log _{2} x}{\log _{2} \frac{1}{2}}=-\log _{2} x
\end{aligned}
$$
the original equation is equivalent to the equation
$$
2 \log ... | 8,\frac{1}{2},\sqrt{2},1,5,\frac{1}{25} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,556 |
Example 15. Solve the equation
$$
\log _{1 / 3}\left(x^{2}+3 x-4\right)=\log _{1 / 3}(2 x+2)
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}+3 x-4>0 \\
2 x+2>0
\end{array}\right.
$$
we find the domain of definition (ODZ) of equation (10): $x>1$.
Potentiating equation (10), we obtain the equation
$$
x^{2}+3 x-4=2 x+2
$$
which is a consequence of it.
Equation (11) has two root... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,557 |
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