problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
3.362. $\frac{6 \sin \alpha-7 \cos \alpha+1}{8 \sin \alpha+9 \cos \alpha-1}$, if $\operatorname{tg} \frac{\alpha}{2}=4$. | Solution.
$$
\begin{aligned}
& \frac{6 \sin \alpha-7 \cos \alpha+1}{8 \sin \alpha+9 \cos \alpha-1}=\left[\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan^{2} \frac{x}{2}} ; \cos x=\frac{1-\tan^{2} \frac{x}{2}}{1+\tan^{2} \frac{x}{2}}\right]= \\
& =\frac{\frac{12 \tan \frac{\alpha}{2}}{1+\tan^{2} \frac{\alpha}{2}}-\frac{7\left(... | -\frac{85}{44} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,143 |
3.363. $\operatorname{tg}\left(\frac{5 \pi}{4}+x\right)+\operatorname{tg}\left(\frac{5 \pi}{4}-x\right)$, if $\operatorname{tg}\left(\frac{3 \pi}{2}+x\right)=\frac{3}{4}$. | Solution.
$$
\begin{aligned}
& \tan\left(\frac{5 \pi}{4}+x\right)+\tan\left(\frac{5 \pi}{4}-x\right)=\tan\left(\left(\frac{3 \pi}{2}+x\right)-\frac{\pi}{4}\right)-\tan\left(x-\frac{5 \pi}{4}\right)= \\
& =\tan\left(\left(\frac{3 \pi}{2}+x\right)-\frac{\pi}{4}\right)-\tan\left(\left(\frac{3 \pi}{2}+x\right)-\frac{11 \p... | -\frac{50}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,144 |
3.364. $\sin \frac{\alpha+\beta}{2}$ and $\cos \frac{\alpha+\beta}{2}$, if $\sin \alpha+\sin \beta=-\frac{21}{65}$;
$$
\cos \alpha+\cos \beta=-\frac{27}{65} ; \frac{5}{2} \pi<\alpha<3 \pi \text { and }-\frac{\pi}{2}<\beta<0
$$ | Solution.
From the condition we have:
$$
\begin{aligned}
& \left\{\begin{array}{l}
\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=-\frac{21}{65} \\
\cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=-\frac{27}{65}
\end{array} \Rightarrow\right. \\
& \Rig... | \sin\frac{\alpha+\beta}{2}=-\frac{7}{\sqrt{130}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,145 |
3.365. $\cos \frac{\alpha-\beta}{2}$, if $\sin \alpha+\sin \beta=-\frac{27}{65} ; \operatorname{tg} \frac{\alpha+\beta}{2}=\frac{7}{9} ; \frac{5}{2} \pi<\alpha<3 \pi$ and $-\frac{\pi}{2}<\beta<0$. | ## Solution.
$$
\begin{aligned}
& \sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} = -\frac{27}{65} \\
& \operatorname{tg} \frac{\alpha + \beta}{2} = \frac{7}{9} \Rightarrow \frac{\sin \frac{\alpha + \beta}{2}}{\cos \frac{\alpha + \beta}{2}} = \frac{7}{9} \Rightarrow \frac{\sin^... | \frac{27}{7\sqrt{130}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,146 |
3.366. $\sin ^{3} \alpha-\cos ^{3} \alpha$, if $\sin \alpha-\cos \alpha=n$. | Solution.
$\sin ^{3} \alpha-\cos ^{3} \alpha=(\sin \alpha-\cos \alpha)\left(\sin ^{2} \alpha+\sin \alpha \cos \alpha+\cos ^{2} \alpha\right)=$
$=[\sin \alpha-\cos \alpha=n]=n(1+\sin \alpha \cos \alpha)=\left[\sin \alpha \cos \alpha=\frac{1-n^{2}}{2}\right]=$
$=n\left(1+\frac{1-n^{2}}{2}\right)=\frac{n\left(2+1-n^{2}... | \frac{3n-n^{3}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,147 |
3.367. $\frac{2 \sin 2 \alpha-3 \cos 2 \alpha}{4 \sin 2 \alpha+5 \cos 2 \alpha}$, if $\operatorname{tg} \alpha=3$. | Solution.
$\frac{2 \sin 2 \alpha-3 \cos 2 \alpha}{4 \sin 2 \alpha+5 \cos 2 \alpha}=\left[\sin x=\frac{2 \operatorname{tg} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}} ; \cos x=\frac{1-\operatorname{tg}^{2} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}}\right]=$
$=\frac{\frac{4 \operatorname{tg} \alpha}{1+\ope... | -\frac{9}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,148 |
3.368. $\sin A+\sin B+\sin C=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$. | ## Solution.
$(\sin A+\sin B)+\sin C=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}+\sin C=$
$=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}+\sin \left(180^{\circ}-(A+B)\right)=$
$=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}+\sin (A+B)=$
$=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}+2 \sin \frac{A+B}{2} \cos \frac{A+B}{2}=$
$=2 \sin \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,149 |
3.369. $\frac{\sin A+\sin B+\sin C}{\sin A+\sin B-\sin C}=\operatorname{ctg} \frac{A}{2} \operatorname{ctg} \frac{B}{2}$. | Solution.
$\frac{\sin A+\sin B+\sin C}{\sin A+\sin B-\sin C}=\left[\sin A+\sin B+\sin C=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\right.$
(see No. 3.368), $\left.\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}\right]=$
$$
\begin{aligned}
& =\frac{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}... | \operatorname{ctg}\frac{A}{2}\operatorname{ctg}\frac{B}{2} | Algebra | proof | Yes | Yes | olympiads | false | 48,150 |
3.370. $\sin 2 A+\sin 2 B+\sin 2 C=4 \sin A \sin B \sin C$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
3.370. $\sin 2 A+\sin 2 B+\sin 2 C=4 \sin A \sin B \sin C$. | ## Solution.
$$
\begin{aligned}
& (\sin 2A + \sin 2B) + \sin 2C = \left[\sin x + \sin y = 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}\right] = \\
& = 2 \sin (A+B) \cos (A-B) + \sin 2C = \\
& = 2 \sin (A+B) \cos (A-B) + \sin 2\left(180^\circ - (A+B)\right) = \\
& = 2 \sin (A+B) \cos (A-B) + \sin \left(360^\circ - 2(A+B)\ri... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,151 |
3.371. $\frac{\sin C}{\cos A \cos B}=\tan A+\tan B$. | Solution.
$\frac{\sin C}{\cos A \cos B}=\frac{\sin \left(180^{\circ}-(A+B)\right)}{\cos A \cos B}=\frac{\sin (A+B)}{\cos A \cos B}=$
$=\frac{\sin A \cos B+\cos A \sin B}{\cos A \cos B}=\frac{\sin A \cos B}{\cos A \cos B}+\frac{\cos A \sin B}{\cos A \cos B}=\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}=$
$=\tan A+\tan ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,152 |
3.372. $\sin 3 A+\sin 3 B+\sin 3 C=-4 \cos \frac{3}{2} A \cos \frac{3}{2} B \cos \frac{3}{2} C$. | ## Solution.
$\sin 3 A+\sin 3 B+\sin 3 C=\left[\sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}\right]=$
$=2 \sin \frac{3 A+3 B}{2} \cos \frac{3 A-3 B}{2}+\sin 3 C=$
$=2 \sin \frac{3 A+3 B}{2} \cos \frac{3 A-3 B}{2}+\sin (540^{\circ}-3(A+B))=$
$=2 \sin \frac{3 A+3 B}{2} \cos \frac{3 A-3 B}{2}+\sin (3 A+3 B)=$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,153 |
3.373. $\sin 4 A+\sin 4 B+\sin 4 C=-4 \sin 2 A \sin 2 B \sin 2 C$. | Solution.
$$
\begin{aligned}
& \sin 4 A+\sin 4 B+\sin 4 C=\left[\sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}\right]= \\
& =2 \sin (2 A+2 B) \cos (2 A-2 B)+\sin 4 C= \\
& =2 \sin (2 A+2 B) \cos (2 A-2 B)+\sin 4\left(180^{\circ}-(A+B)\right)= \\
& =2 \sin (2 A+2 B) \cos (2 A-2 B)+\sin \left(720^{\circ}-4(A+B)\r... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,154 |
3.374. $\operatorname{tg} \frac{A}{2} \operatorname{tg} \frac{B}{2}+\operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2}+\operatorname{tg} \frac{C}{2} \operatorname{tg} \frac{A}{2}=1$.
3.374. $\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$. | Solution.
$$
\begin{aligned}
& \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = \tan \frac{A}{2} \tan \frac{B}{2} + \left( \tan \frac{B}{2} + \tan \frac{A}{2} \right) \tan \frac{C}{2} = \\
& = \tan \frac{A}{2} \tan \frac{B}{2} + \left( \tan \frac{B}{2} + \tan ... | 1 | Geometry | proof | Yes | Yes | olympiads | false | 48,155 |
3.375. Find $\operatorname{tg} \frac{x}{2}$, if it is known that $\sin x+\cos x=\frac{1}{5}$. | Solution.
Let $\sin x + \cos x = \frac{1}{5}$.
Since $\sin \alpha = \frac{2 \operatorname{tg} \frac{\alpha}{2}}{1 + \operatorname{tg}^{2} \frac{\alpha}{2}} ; \cos \alpha = \frac{1 - \operatorname{tg}^{2} \frac{\alpha}{2}}{1 + \operatorname{tg}^{2} \frac{\alpha}{2}}, \alpha \neq (2 n + 1) \pi, n \in Z$, then
we have ... | \operatorname{tg}\frac{x}{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,156 |
3.376. Knowing that $\operatorname{tg} \frac{\alpha}{2}=m$, find $\frac{1-2 \sin ^{2} \frac{\alpha}{2}}{1+\sin \alpha}$. | Solution.
$$
\begin{aligned}
& \frac{1-2 \sin ^{2} \frac{\alpha}{2}}{1+\sin \alpha}=\left[1-2 \sin ^{2} x=\cos 2 x\right]=\frac{\cos \alpha}{1+\sin \alpha}= \\
& =\left[\cos x=\frac{1-\operatorname{tg}^{2} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}} ; \sin x=\frac{2 \operatorname{tg} \frac{x}{2}}{1+\operatorname... | \frac{1-}{1+} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,157 |
3.377. Find the value of the expression $\frac{1+\cos 2 \alpha}{\operatorname{ctg} \frac{\alpha}{2}-\operatorname{tg} \frac{\alpha}{2}}$, given that $\sin \alpha+\cos \alpha=m$. | Solution.
$$
\begin{aligned}
& \frac{1+\cos 2 \alpha}{\cot \frac{\alpha}{2}-\tan \frac{\alpha}{2}}=\frac{1+\cos ^{2} \alpha-\sin ^{2} \alpha}{\frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}-\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}}=\frac{2 \cos ^{2} \alpha}{\frac{\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \fr... | \frac{^{2}-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,158 |
3.380. It is known that $\operatorname{tg} \alpha=\frac{p}{q}$. Find $\sin 2 \alpha, \cos 2 \alpha$ and $\operatorname{tg} 2 \alpha$. | ## Solution.
By the formulas of universal trigonometric substitution we have
$$
\begin{aligned}
& \sin 2 \alpha=\frac{2 \tan \alpha}{1+\tan^{2} \alpha}, \alpha \neq(2 n+1) \pi ; \cos 2 \alpha=\frac{1-\tan^{2} \alpha}{1+\tan^{2} \alpha}, \alpha \neq(2 n+1) \pi ; \\
& \tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan^{2} \alph... | \sin2\alpha=\frac{2pq}{p^{2}+q^{2}};\cos2\alpha=\frac{q^{2}-p^{2}}{q^{2}+p^{2}};\tan2\alpha=\frac{2pq}{q^{2}-p^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,160 |
3.381. Find $\cos 2 \alpha$, if it is known that $2 \operatorname{ctg}^{2} \alpha+7 \operatorname{ctg} \alpha+3=0$ and the number $\alpha$ satisfies the inequalities: a) $\frac{3 \pi}{2}<\alpha<\frac{7 \pi}{4}$; b) $\frac{7 \pi}{4}<\alpha<2 \pi$. | ## Solution.
Solving the equation $2 \operatorname{ctg}^{2} \alpha+7 \operatorname{ctg} \alpha+3=0$ as a quadratic equation in $\operatorname{ctg} \alpha$, we have $(\operatorname{ctg} \alpha)_{1,2}=\frac{-7 \pm \sqrt{49-24}}{4}=\frac{-7 \pm 5}{4}$, from which $\operatorname{ctg} \alpha=-3$, $\operatorname{tg} \alpha=... | -\frac{3}{5};\frac{4}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,161 |
3.382. Find $\sin 2 \alpha$, if it is known that $2 \operatorname{tg}^{2} \alpha-7 \operatorname{tg} \alpha+3=0$ and the number $\alpha$ satisfies the inequalities: a) $\pi<\alpha<\frac{5 \pi}{4}$; b) $\frac{5 \pi}{4}<\alpha<\frac{3 \pi}{2}$. | ## Solution.
Solving the equation $2 \operatorname{tg}^{2} \alpha-7 \operatorname{tg} \alpha+3=0$ as a quadratic equation in $\operatorname{tg} \alpha$, we get $(\operatorname{tg} \alpha)_{1,2}=\frac{7 \pm \sqrt{49-24}}{4}=\frac{7 \pm 5}{4}, \operatorname{tg} \alpha=\frac{1}{2}$ or $\operatorname{tg} \alpha=3$.
In ca... | \frac{4}{5};\frac{3}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,162 |
3.384. Prove that the expression $\frac{1-2 \sin ^{2}\left(\alpha-\frac{3}{2} \pi\right)+\sqrt{3} \cos \left(2 \alpha+\frac{3}{2} \pi\right)}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}$
does not depend on $\alpha$, where $\alpha \neq \frac{\pi n}{2}+\frac{\pi}{12}$. | ## Solution.
$$
\begin{aligned}
& \frac{1-2 \sin ^{2}\left(\alpha-\frac{3}{2} \pi\right)+\sqrt{3} \cos \left(2 \alpha+\frac{3}{2} \pi\right)}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}= \\
& =\frac{1-2\left(-\sin \left(\frac{3}{2} \pi-\alpha\right)\right)^{2}+\sqrt{3} \cos \left(\frac{3}{2} \pi+2 \alpha\right)}{\sin \l... | -2 | Algebra | proof | Yes | Yes | olympiads | false | 48,164 |
3.385. Prove that $\operatorname{tg} \frac{\alpha}{2}+\operatorname{tg} \frac{\beta}{2}+\operatorname{tg} \frac{\gamma}{2}=\operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\gamma}{2}$, if $\alpha+\beta+\gamma=2 \pi$. | ## Solution.
$$
\begin{aligned}
& \left(\operatorname{tg} \frac{\alpha}{2}+\operatorname{tg} \frac{\beta}{2}\right)+\operatorname{tg} \frac{\gamma}{2}=\left[\operatorname{tg} x+\operatorname{tg} y=\frac{\sin (x+y)}{\cos x \cos y}, x, y \neq \frac{\pi}{2}+\pi n, n \in Z\right]= \\
& =\frac{\sin \frac{\alpha+\beta}{2}}{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,165 |
3.387. Prove that the expression $\frac{1-\cos ^{4}\left(\alpha-\frac{3}{2} \pi\right)-\sin ^{4}\left(\alpha+\frac{3}{2} \pi\right)}{\sin ^{6} \alpha+\cos ^{6} \alpha-1}$ does not depend on $\alpha$, if $\alpha \neq \frac{\pi n}{2}$. | Solution.
$$
\begin{aligned}
& \frac{1-\cos ^{4}\left(\alpha-\frac{3}{2} \pi\right)-\sin ^{4}\left(\alpha+\frac{3}{2} \pi\right)}{\sin ^{6} \alpha+\cos ^{6} \alpha-1}= \\
& =\frac{1-\left(\cos \left(\frac{3}{2} \pi-\alpha\right)\right)^{4}-\left(\sin \left(\frac{3}{2} \pi+\alpha\right)\right)^{4}}{\left(\sin ^{2} \alp... | -\frac{2}{3} | Algebra | proof | Yes | Yes | olympiads | false | 48,166 |
3.388. Prove that the expression $\sin \left(250^{\circ}+\alpha\right) \cos \left(200^{\circ}-\alpha\right)-\cos 240^{\circ} \times$ $\cos \left(220^{\circ}-2 \alpha\right)$ does not depend on $\alpha$. | ## Solution.
$$
\begin{aligned}
& \sin \left(250^{\circ}+\alpha\right) \cos \left(200^{\circ}-\alpha\right)-\cos 240^{\circ} \cos \left(220^{\circ}-2 \alpha\right)= \\
& =\sin \left(270^{\circ}+\left(\alpha-20^{\circ}\right)\right) \cos \left(180^{\circ}-\left(\alpha-20^{\circ}\right)\right)- \\
& -\cos \left(270^{\ci... | \frac{1}{2} | Algebra | proof | Yes | Yes | olympiads | false | 48,167 |
3.389. Prove that the expression $\cos ^{2} \alpha+\cos ^{2} \varphi+\cos ^{2}(\alpha+\varphi)-$ $2 \cos \alpha \cos \varphi \cos (\alpha+\varphi)$ does not depend on either $\alpha$ or $\varphi$. | ## Solution.
$\cos ^{2} \alpha+\cos ^{2} \varphi+\cos ^{2}(\alpha+\varphi)-2 \cos \alpha \cos \varphi \cos (\alpha+\varphi)=$ $=\cos ^{2} \alpha+\cos ^{2} \varphi+(\cos (\alpha+\varphi))^{2}-2 \cos \alpha \cos \varphi \times$
$x(\cos \alpha \cos \varphi-\sin \alpha \sin \varphi)=\cos ^{2} \alpha+\cos ^{2} \varphi+\cos... | 1 | Algebra | proof | Yes | Yes | olympiads | false | 48,168 |
3.390. Derive the formula $\cos (n+1) \alpha=2 \cos \alpha \cos n \alpha-\cos (n-1) \alpha$, where $n$ is any real number, and use it to express $\cos 3 \alpha$ and $\cos 4 \alpha$ as polynomials in $\cos \alpha$. | Solution.
$\cos (n+1) \alpha=\cos (n \alpha+\alpha)=[\cos (x+y)=\cos x \cos y-\sin x \sin y]=$
$=\cos n \alpha \cos \alpha-\sin n \alpha \sin \alpha=2 \cos n \alpha \cos \alpha-\cos n \alpha \cos \alpha-\sin n \alpha \sin \alpha:=$
$=2 \cos n \alpha \cos \alpha-(\cos n \alpha \cos \alpha+\sin n \alpha \sin \alpha)=$... | \cos3\alpha=4\cos^{3}\alpha-3\cos\alpha;\cos4\alpha=8\cos^{4}\alpha-8\cos^{2}\alpha+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,169 |
3.391. Prove that $4 \sin \left(30^{\circ}+\frac{\alpha}{2}\right) \sin \left(30^{\circ}-\frac{\alpha}{2}\right)=\frac{\cos \frac{3 \alpha}{2}}{\cos \frac{\alpha}{2}}$. | ## Solution.
$$
\begin{aligned}
& 4 \sin \left(30^{\circ}+\frac{\alpha}{2}\right) \sin \left(30^{\circ}-\frac{\alpha}{2}\right)=\left[\sin x \sin y=\frac{1}{2}(\cos (x-y)-\cos (x+y))\right]= \\
& =2\left(\cos \alpha-\cos 60^{\circ}\right)=2\left(\cos \alpha-\frac{1}{2}\right)=2 \cos \alpha-1=2\left(2 \cos ^{2} \frac{\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,170 |
3.393. Show that if $p$ is constant, then the function
$f(a)=\frac{p \cos ^{3} \alpha-\cos 3 \alpha}{\cos \alpha}+\frac{p \sin ^{3} \alpha+\sin 3 \alpha}{\sin \alpha}$ is also constant. | ## Solution.
Using the formulas $\cos 3 x=4 \cos ^{3} x-3 \cos x$ and $\sin 3 x=3 \sin x-4 \sin ^{3} x$, we have
$$
\begin{aligned}
& f(a)=\frac{p \cos ^{3} \alpha-\left(4 \cos ^{3} \alpha-3 \cos \alpha\right)}{\cos \alpha}+ \\
& +\frac{p \sin ^{3} \alpha+3 \sin \alpha-4 \sin ^{3} \alpha}{\sin \alpha}= \\
& =\frac{p ... | p+2 | Algebra | proof | Yes | Yes | olympiads | false | 48,172 |
3.394. Given the function $f(x)=\cos ^{4} x+\sin ^{4} x$. Find $f(\alpha)$, if it is known that $\sin 2 \alpha=\frac{2}{3}$. | ## Solution.
$$
\begin{aligned}
& f(x)=\cos ^{4} x+\sin ^{4} x=\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-2 \cos ^{2} x \sin ^{2} x= \\
& =1-2 \cos ^{2} x \sin ^{2} x=1-\frac{1}{2}\left(4 \cos ^{2} x \sin ^{2} x\right)=1-\frac{1}{2} \sin ^{2} 2 x \\
& f(\alpha)=1-\frac{1}{2} \sin ^{2} 2 \alpha=1-\frac{1}{2}(\sin 2 \alph... | \frac{7}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,173 |
3.395. Prove that if $\alpha+\beta=60^{\circ}(\alpha>0, \beta>0)$, then $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta \leq \frac{1}{3}$. | ## Solution.
$$
\begin{aligned}
& \operatorname{tg} \alpha \cdot \operatorname{tg} \beta \leq \frac{1}{3} \Leftrightarrow \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} \leq \frac{1}{3} \Leftrightarrow \\
& \Leftrightarrow \frac{\frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta))}{\frac{1}{2}(\cos (\alpha-\bet... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 48,174 |
4.036. The sum of the first three terms of a geometric progression is 21, and the sum of their squares is 189. Find the first term and the common ratio of this progression. | ## Solution.
From the condition we have
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ b _ { 1 } + b _ { 2 } + b _ { 3 } = 21, } \\
{ b _ { 1 } ^ { 2 } + b _ { 2 } ^ { 2 } + b _ { 3 } ^ { 2 } = 189, }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}+b_{1} q+b_{1} q^{2}=21, \\
b_{1}^{2}+b_{1}^{2} q^{2}+b_... | 12,\frac{1}{2};3,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,175 |
4.037. Prove that any term of an arithmetic progression, starting from the second, is the arithmetic mean of any two terms equidistant from it. | ## Solution.
Let $a_{k}$ be any term of the arithmetic progression, then $a_{k-p}, a_{k+p}$ are two terms equally distant from it.
Using the formula for the general term, we find
$$
\begin{aligned}
& a_{k-p}=a_{1}+d(k-p-1) \\
& a_{k+p}=a_{1}+d(k+p-1)
\end{aligned}
$$
Adding these equations, we get
$$
a_{k-p}+a_{k+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,176 |
4.040. Three numbers form a geometric progression. If 4 is subtracted from the third number, the numbers will form an arithmetic progression. If 1 is then subtracted from the second and third terms of the resulting arithmetic progression, a geometric progression will again be obtained. Find these numbers. | Solution.
Let $b_{1}, b_{1} q, b_{1} q^{2}$ be the given numbers, then
$b_{1}, b_{1} q, b_{1} q^{2}$ - terms of a geometric progression,
$b_{1}, b_{1} q, b_{1} q^{2}-4$ - terms of an arithmetic progression,
$b_{1}, b_{1} q-1, b_{1} q^{2}-5$ - terms of a geometric progression.
Using the properties of the terms of an... | \frac{1}{9},\frac{7}{9},\frac{49}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,179 |
4.041. Find the positive integer $n$ from the equation
$$
(3+6+9+\ldots+3(n-1))+\left(4+5.5+7+\ldots+\frac{8+3 n}{2}\right)=137
$$ | ## Solution.
In the first parentheses, there is the sum of the terms of an arithmetic progression $S_{k}$ where $a_{1}=3, d=3, a_{k}=3(n-1), k=\frac{a_{k}-a_{1}}{d}+1=\frac{3 n-3-3}{3}+1=n-1$; in the second parentheses, there is the sum of the terms of an arithmetic progression where $b_{1}=4, d=1.5, a_{m}=\frac{8+3 n... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,180 |
4.043. The sum of an infinite geometric progression with a common ratio $|q|<1$ is 4, and the sum of the cubes of its terms is 192. Find the first term and the common ratio of the progression. | ## Solution.
From the condition we have
$$
\begin{aligned}
& \left\{\begin{array}{l}
b_{1}+b_{2}+b_{3}+\ldots=4, \\
b_{1}^{3}+b_{2}^{3}+b_{3}^{3}+\ldots=192 \\
|q|<1
\end{array}, \Leftrightarrow\left\{\begin{array}{l}
b_{1}+b_{1} q+b_{1} q^{2}+\ldots=4, \\
b_{1}^{3}+b_{1}^{3} q^{3}+b_{1}^{3} q^{6} \ldots=192,
\end{ar... | 6,-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,182 |
4.044. Find four numbers, the first three of which form a geometric progression, and the last three form an arithmetic progression. The sum of the extreme numbers is 21, and the sum of the middle numbers is 18. | ## Solution.
Let $b_{1}, b_{1} q, b_{1} q^{2}, b_{1} q^{2}+d$ be the given numbers.
From the condition, we have:
$b_{1}, b_{1} q, b_{1} q^{2}$ are terms of a geometric progression,
$b_{1}, b_{1} q+d, b_{1} q+2 d$ are terms of an arithmetic progression,
$b_{1}+b_{1} q+2 d=21, b_{1} q+b_{1} q+d=18$.
Since $b_{3}=b_... | 3,6,12,18;18.75;11.25;6.75;2.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,183 |
4.045. The sum of the first three terms of a geometric progression is 91. If 25, 27, and 1 are added to these terms respectively, the resulting three numbers form an arithmetic progression. Find the seventh term of the geometric progression. | ## Solution.
From the condition, we have:
$b_{1}, b_{1} q, b_{1} q^{2}$ - terms of a geometric progression,
$b_{1}+25, b_{1} q+27, b_{1} q^{2}$ - terms of an arithmetic progression, then we get
$$
\left\{\begin{array} { l }
{ b _ { 1 } + b _ { 1 } q + b _ { 1 } q ^ { 2 } = 9 1 , } \\
{ 2 ( b _ { 1 } q + 2 7 ) = b ... | 5103 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,184 |
4.046. Three numbers form a geometric progression. If the second number is increased by 2, the progression becomes arithmetic, and if after that the last number is increased by 9, the progression becomes geometric again. Find these numbers. | ## Solution.
Let $b_{1}, b_{2}, b_{3}$ be terms of a geometric progression, $b_{1}, b_{2}+2, b_{3}--$ be terms of an arithmetic progression, and $b_{1}, b_{2}+2, b_{3}+9$ be terms of a geometric progression. Then $b_{1}, b_{1} q+2, b_{1} q^{2}+9$ are terms of a geometric progression.
We have:
$$
\begin{aligned}
& \l... | 4,8,16;\frac{4}{25},-\frac{16}{25},\frac{64}{25} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,185 |
4.047. Find three numbers that form a geometric progression, given that their product is 64, and their arithmetic mean is $14 / 3$. | ## Solution.
From the condition we have:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ b _ { 1 } \cdot b _ { 2 } \cdot b _ { 3 } = 6 4 } \\
{ \frac { b _ { 1 } + b _ { 2 } + b _ { 3 } } { 3 } \frac { 1 4 } { 3 } } \\
{ \{ \begin{array} { l }
{ b _ { 1 } \cdot b _ { 1 } q \cdot b _ { 1 } q ^ { 2 } = 6 4 , } \\
{... | 2,4,88,4,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,186 |
4.048. Prove that any term of a sign-positive geometric progression, starting from the second, is equal to the mean proportional between any terms equally distant from it. | Solution.
We have $b_{k-p}=b_{1} q^{k-p-1}, b_{k+p}=b_{1} q^{k+p-1}$.
Multiplying these equalities, we get $b_{k-p} \cdot b_{k+p}=b_{1}^{2} q^{2(k-1)}$ or $\left(b_{1} q^{k-1}\right)^{2}=b_{k-p} \cdot b_{k+p}$, from which $b_{1} q^{k-1}=\sqrt{b_{k-p} \cdot b_{k+p}}$, i.e., $b_{k}=\sqrt{b_{k-p} \cdot b_{k+p}}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,187 |
4.050. Find the sum of all three-digit numbers divisible by 7. | Solution.
We have:
$$
\left\{\begin{array}{l}
a_{1}=105 \\
a_{n}=994, \\
d=7
\end{array}\right.
$$
Answer: 70336. | 70336 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 48,188 |
4.052. Given two infinite geometric progressions with a common ratio $|q|<1$, differing only in the sign of their common ratios. Their sums are respectively equal to $S_{1}$ and $S_{2}$. Find the sum of the infinite geometric progression formed by the squares of the terms of any of the given progressions. | Solution.
From the condition, we have $S_{1}=\frac{b_{1}}{1-q}, S_{2}=\frac{b_{1}}{1+q},|q|<1$. Further,
$$
b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+\ldots=b_{1}^{2}+b_{1}^{2} q^{2}+b_{1}^{2} q^{4}+\ldots=b_{1}^{2}\left(1+q^{2}+q^{4}+\ldots\right)
$$
The expression in parentheses is the sum of the terms of an infinitely decrea... | S_{1}\cdotS_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,190 |
4.053. Let $a_{1}, a_{2}, \ldots a_{n}$ be the consecutive terms of a geometric progression, $S_{n}$ - the sum of its first $n$ terms. Prove that
$$
S_{n}=a_{1} a_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}\right)
$$ | ## Solution.
From the condition $S_{n}=\frac{a_{1}\left(1-q^{n}\right)}{1-q}$.
## Further, we have:
$$
\begin{aligned}
& a_{1} a_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}\right)=a_{1} \cdot a_{1} q^{n-1}\left(\frac{1}{a_{1}}+\frac{1}{a_{1} q}+\frac{1}{a_{1} q^{2}}+\ldots+\frac{1}{a_{1} q^{n-1}}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,191 |
4.054. Prove that if the numbers $a, b$ and $c$ form an arithmetic progression, then the numbers $a^{2}+a b+b^{2}, a^{2}+a c+c^{2}$ and $b^{2}+b c+c^{2}$ in the given order also form an arithmetic progression. | Solution.
We have $b=a+d, c=a+2 d$.
Assume that $a^{2}+a(a+d)+(a+d)^{2}, a^{2}+a(a+2 d)+(a+2 d)^{2}$, $(a+d)^{2}+(a+d)+(a+2 d)+(a+2 d)^{2}$ are terms of an arithmetic progression.
Then we get
$$
\begin{aligned}
& 2\left(a^{2}+a(a+2 d)+(a+2 d)^{2}\right)=\left(a^{2}+a(a+d)+(a+d)^{2}\right)+ \\
& +\left((a+d)^{2}+(a+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,192 |
4.055. The first term of a certain infinite geometric progression with a common ratio $|q|<1$ is 1, and its sum is $S$. A new infinite geometric progression is formed from the squares of the terms of this progression. Find its sum. | ## Solution.
From the condition, we have $b_{1}=1,|q|<1$. Therefore, $\frac{1}{1-q}=S, 1-q=\frac{1}{S}$, $q=1-\frac{1}{S}=\frac{S-1}{S}$.
Then $b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+\ldots=b_{1}^{2}+b_{1}^{2} q^{2}+b_{1}^{2} q^{4}+\ldots=$
$=b_{1}^{2}\left(1+q^{2}+q^{4}+\ldots\right)=b_{1}^{2} \cdot \frac{1}{1-q^{2}}=\frac{... | \frac{S^{2}}{2S-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,193 |
4.056. Find the fifth term of an increasing geometric progression, given that its first term is equal to $7-3 \sqrt{5}$ and that each of its terms, starting from the second, is equal to the difference of the two adjacent terms. | Solution.
We have:
$$
\left\{\begin{array}{l}
b_{1}=7-3 \sqrt{5} \\
b_{2}=b_{3}-b_{1}, \\
|q|>1
\end{array}\right.
$$
$q_{1}=\frac{1-\sqrt{5}}{2}$ does not fit, as $\left|q_{1}\right|<1 ; q_{2}=\frac{1+\sqrt{5}}{2}$.
From this,
$$
b_{5}=b_{1} q^{4}=\frac{(7-3 \sqrt{5})(1+\sqrt{5})^{4}}{16}=\frac{(17-3 \sqrt{5}) 8(... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,194 |
4.057. In an arithmetic progression, the sum of its first $m$ terms is equal to the sum of its first $n$ terms $(m \neq n)$. Prove that in this case, the sum of its first $m+n$ terms is zero. | Solution.
From the condition $a_{1}, a_{2}, a_{3}, \ldots, a_{m}, a_{m+1}, a_{m+2}, \ldots, a_{n}$, where $m \neq n, S_{m}=S_{n}$.
$$
\begin{aligned}
& S_{m}=\frac{2 a_{1}+(m-1) d}{2} \cdot m, S_{n}=\frac{a_{1}+a_{n}}{2} \cdot n \\
& \text { Then } \frac{2 a_{1}+(m-1) d}{2} \cdot m=\frac{2 a_{1}+(n-1) d}{2} \cdot n \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,195 |
4.058. It is known that $L, M, N$ are the $l$-th, $m$-th, $n$-th terms of a geometric progression. Show that $L^{m-n} M^{n-l} N^{l-m}=1$. | ## Solution.
Let $b_{1}, b_{2}, \ldots, b_{l}, \ldots, b_{m}, \ldots, b_{n}, \ldots$ be the terms of a geometric progression and let $L=b_{l}, M=b_{m}$ and $N=b_{n}$. We will show that $b_{l}^{m-n} \cdot b_{m}^{n-l} \cdot b_{n}^{l-m}=1$, where $b_{l}=b_{1} q^{l-1}$, $b_{m}=b_{1} q^{m-1}$, and $b_{n}=b_{1} q^{n-1}$.
W... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,196 |
4.060. Show that for any arithmetic progression and for any n, the equality $S_{2 n}=S_{n}+\frac{1}{3} S_{3 n}\left(S_{k}\right.$ - the sum of the first $k$ terms of the progression) holds. | Solution.
We have $S_{n}=\frac{2 a_{1}+d(n-1)}{2} n$,
$$
\begin{aligned}
& \frac{1}{3} S_{3 n}=\frac{2 a_{1}+d(3 n-1)}{2 \cdot 3} \cdot 3 n=\frac{2 a_{1}+d(3 n-1)}{2} \cdot n \\
& S_{n}+\frac{1}{3} S_{3 n}=\frac{4 a_{1}+d(4 n-2)}{2} \cdot n=\frac{2 a_{1}+d(2 n-2)}{2} \cdot 2 n=S_{2 n}
\end{aligned}
$$
This completes... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,198 |
4.061. Solve the equation
$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x}+\ldots+\frac{1}{x}=3
$$
where $x$ is a positive integer. | ## Solution.
Multiplying both sides of the equation by $x$, we have
$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$
The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1, a_{n}=1$,
$$
n=\frac{a_{n}-a_{1}}{d}+1=\frac{1-x+1}{-1}+1=x-1 . \text { Since } S_{n}=\frac{a_{1}+a_{n... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,199 |
4.062. Represent the number 180 as the sum of four terms so that they form a geometric progression, in which the third term is 36 more than the first. | Solution.
From the condition
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ b _ { 1 } + b _ { 2 } + b _ { 3 } + b _ { 4 } = 1 8 0 , } \\
{ b _ { 3 } = b _ { 1 } + 3 6 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}+b_{1} q+b_{1} q^{2}+b_{1} q^{3}=180, \\
b_{1} q^{2}=b_{1}+36,
\end{array} \Leftrighta... | 12+24+48+96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,200 |
4.063. Given two geometric progressions consisting of the same number of terms. The first term and the common ratio of the first progression are 20 and $3 / 4$, respectively, while the first term and the common ratio of the second progression are 4 and $2 / 3$, respectively. If the terms of these progressions with the ... | ## Solution.
Let $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ be the terms of two geometric progressions, where $a_{1}=20, q_{1}=\frac{3}{4}; b_{1}=4, q_{2}=\frac{2}{3}$ and $a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+\ldots=$ $=158.75$. We find $a_{2}=a_{1} q_{1}=20 \cdot \frac{3}{4}=15, a_{3}=a_{1} q_{1}... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,201 |
4.064. Three numbers, of which the third is 12, form a geometric progression. If 12 is replaced by 9, the three numbers form an arithmetic progression. Find these numbers. | ## Solution.
Let $b_{1}, b_{2}, 12$ be terms of a geometric progression, and $\mathrm{a} b_{1}, b_{2}, 9$ be terms of an arithmetic progression. Then
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ b _ { 1 } q ^ { 2 } = 1 2 , } \\
{ b _ { 1 } + 2 d = 9 , } \\
{ b _ { 1 } q = b _ { 1 } + d , }
\end{array} \Leftrigh... | 27,18,12;3,6,12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,202 |
4.065. In a finite geometric progression, the first term $a$, the last term $b$, and the sum $S$ of all its terms are known. Find the sum of the squares of all the terms of this progression. | ## Solution.
Let $b_{1}, b_{2}, b_{3}, \ldots b_{n}$ be the terms of a geometric progression, where $b_{1}=a, b_{n}=b, S_{n}=S$. Then
$$
\begin{aligned}
& b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+\ldots=b_{1}^{2}+b_{1}^{2} q^{2}+b_{1}^{2} q^{4}+\ldots=b_{1}^{2}\left(1+q^{2}+q^{4}+\ldots\right)= \\
& =a^{2}\left(1+q^{2}+q^{4}+\l... | \frac{(+b)S-2}{2S-(+b)}\cdotS | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,203 |
4.066. In a certain geometric progression containing $2 n$ positive terms, the product of the first term and the last is 1000. Find the sum of the decimal logarithms of all the terms of the progression. | ## Solution.
Let $b_{1}, b_{2}, b_{3}, \ldots, b_{n}$ be the terms of a geometric progression, the number of which is $2n$, and $b_{1} \cdot b_{n}=1000$. We have
$$
\begin{aligned}
& X=\lg b_{1}+\lg b_{2}+\lg b_{3}+\ldots+\lg b_{n}=\lg \left(b_{1} \cdot b_{2} \cdot b_{3} \ldots b_{n}\right)= \\
& =\lg \left(\left(b_{... | 3n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,204 |
4.067. The sum of three numbers is $11 / 18$, and the sum of their reciprocals, which form an arithmetic progression, is 18. Find these numbers. | ## Solution.
From the condition, we have $a_{1}+a_{2}+a_{3}=\frac{11}{18}, \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}=18$, where $\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}$ are terms of an arithmetic progression.
Thus,
$$
\begin{aligned}
& \left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}=\frac{11}{18}, \\
\fra... | \frac{1}{9},\frac{1}{6},\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,205 |
6.136. $\frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
6.136. $\frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
(Note: The equation is the same in both languages, so the translation is identical to the original text.) | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-1, \\ x \neq 2 .\end{array}\right.$
After bringing all terms of the equation to a common denominator, we have
$$
\begin{aligned}
& \frac{\left(x^{2}+1\right)(x-2)+\left(x^{2}+2\right)(x+1)+2(x+1)(x-2)}{(x+1)(x-2)}=0 \Leftrightarrow \\
& \Leftrightarro... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,207 |
6.137. $\frac{x}{x+1}+\frac{x+1}{x+2}+\frac{x+2}{x}=\frac{25}{6}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-2, \\ x \neq-1, \\ x \neq 0 .\end{array}\right.$
Bring all terms of the equation to a common denominator:
$$
\frac{6 x^{2}(x+2)+6 x(x+1)^{2}+6(x+1)(x+2)^{2}-25 x(x+1)(x+2)}{6 x(x+1)(x+2)}=0 \Leftrightarrow
$$
$\Leftrightarrow \frac{7 x^{3}+21 x^{2}-4... | x_{1}=1,x_{2,3}=-2\\frac{2\sqrt{7}}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,208 |
6.138. $\left(x^{2}-6 x\right)^{2}-2(x-3)^{2}=81$. | ## Solution.
From the condition $\left(x^{2}-6 x\right)^{2}-2\left(x^{2}-6 x+9\right)-81=0$. Let $x^{2}-6 x=y$.
We have: $y^{2}-2(y+9)-81=0, y^{2}-2 y-99=0 \Rightarrow y_{1}=-9 ; \quad y_{2}=11$. Then 1) $\left.x^{2}-6 x=-9,(x-3)^{2}=0, x_{1,2}=3 ; 2\right) \quad x^{2}-6 x=11, x^{2}-6 x-11=0$, $x_{3,4}=3 \pm \sqrt{20... | x_{1,2}=3;x_{3,4}=3\2\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,209 |
6.139. $(x+1)^{5}+(x-1)^{5}=32 x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.139. $(x+1)^{5}+(x-1)^{5}=32 x$. | ## Solution.
Since $a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2} b+a^{n-3} b^{2}-a^{n-4} b^{3}+\ldots+b^{n-1}\right)$, then $(x+1+x-1)\left((x+1)^{4}-(x+1)^{3}(x-1)+(x+1)^{2}(x-1)^{2}-(x+1)(x-1)^{3}+\right.$ $\left.+(x-1)^{4}\right)=32 x, \quad 2 x\left((x+1)^{4}+(x-1)^{4}-(x+1)^{3}(x-1)-(x+1)(x-1)^{3}+\right.$ $\left.+(x+1... | x_{1}=0,x_{2}=-1,x_{3}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,210 |
6.140. $\frac{z^{2}-z}{z^{2}-z+1}-\frac{z^{2}-z+2}{z^{2}-z-2}=1$. | Solution.
Domain of definition: $z^{2}-z-2 \neq 0 \Leftrightarrow\left\{\begin{array}{l}z \neq-1, \\ z \neq 2 .\end{array}\right.$
Let $z^{2}-z=y$. We have:
$\frac{y}{y+1}-\frac{y+2}{y-2}-1=0 \Leftrightarrow \frac{y(y-2)-(y+2)(y+1)-(y+1)(y-2)}{(y+1)(y-2)}=0 \Leftrightarrow$ $\Leftrightarrow \frac{y(y+4)}{(y+1)(y-2)}... | z_{1}=0,z_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,211 |
6.141. $\frac{24}{x^{2}+2 x-8}-\frac{15}{x^{2}+2 x-3}=2$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x^{2}+2 x-8 \neq 0, \\ x^{2}+2 x-3 \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \neq-4, \\ x \neq 2, \\ x \neq-3, \\ x \neq 1 .\end{array}\right.\right.$
Let $x^{2}+2 x=y$.
We have $\quad \frac{24}{y-8}-\frac{15}{y-3}-2=0 \Leftrightarrow \frac{y... | x_{1}=0,x_{2}=-2,x_{3,4}=\frac{-2\\sqrt{66}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,212 |
6.142. $x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c=0$.
6.142. $x^{3}-(a+b+c) x^{2}+(ab+ac+bc) x-abc=0$. | Solution.
By verification, we find that $x_{1}=a$, since $a^{3}-a^{2}-a^{2} b-a^{2} c+ a^{2} b+a^{2} c+a b c-a b c=0$. We divide the left side of the equation by $x-a$:
$\frac{x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c}{x-a}=x^{2}-(b+c) x+b c$.
This equation can be represented as:
$$
(x-a)\left(x^{2}-(b+c) x+b c\rig... | x_{1}=,x_{2}=,x_{3}=b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,213 |
6.143. $\frac{1}{x^{2}}+\frac{1}{(x+2)^{2}}=\frac{10}{9}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ x \neq-2 .\end{array}\right.$
From the condition $\left(\frac{1}{x}\right)^{2}+\left(\frac{1}{x+2}\right)^{2}-\frac{10}{9}=0 \Leftrightarrow\left(\frac{x+2-x}{x(x+2)}\right)^{2}+\frac{2}{x(x+2)}-$ $-\frac{10}{9}=0 \Leftrightarrow\left(\frac{2}{x(... | x_{1}=-3,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,214 |
6.144. $\left(x^{2}+2 x\right)^{2}-(x+1)^{2}=55$.
6.144. $\left(x^{2}+2 x\right)^{2}-(x+1)^{2}=55$. | Solution.
From the condition, we have $\left(x^{2}+2 x\right)^{2}-\left(x^{2}+2 x+1\right)=55$. Let $x^{2}+2 x=y$.
The equation in terms of $y$ becomes $y^{2}-y-56=0 \Rightarrow y_{1}=-7, y_{2}=8$.
Then $x^{2}+2 x=-7$ or $x^{2}+2 x=8, x^{2}+2 x+7=0$ or $x^{2}+2 x-8=0$, from which $x_{1}=-4, x_{2}=2$. The first equat... | x_{1}=-4,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,215 |
6.145. $(x+1)^{2}(x+2)+(x-1)^{2}(x-2)=12$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.145. $(x+1)^{2}(x+2)+(x-1)^{2}(x-2)=12$. | ## Solution.
We have:
$$
\begin{aligned}
& \left(x^{2}+2 x+1\right)(x+2)+\left(x^{2}-2 x+1\right)(x-2)-12=0 \Leftrightarrow \\
& \Leftrightarrow x^{3}+2 x^{2}+2 x^{2}+4 x+x+2+x^{3}-2 x^{2}-2 x^{2}+4 x+x-2-12=0 \\
& 2 x^{3}+10 x-12=0, x^{3}+5 x-6=0
\end{aligned}
$$
The last equation can be rewritten as:
$x^{3}+5 x-5... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,216 |
6.146. $\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-1, \\ x \neq-2, \\ x \neq-3, \\ x \neq-4 .\end{array}\right.$
$$
\begin{aligned}
& \text { From the condition } \frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}-1=0 \Leftrightarrow \\
& \Leftrightarrow \frac{(x-1)(x-4)(x-2)(x-3)-(x+1)(x+4)(x+2)(x+3)}{(x+... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,217 |
6.147. $\frac{6}{(x+1)(x+2)}+\frac{8}{(x-1)(x+4)}=1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq \pm 1, \\ x \neq-2, \\ x \neq-4 .\end{array}\right.$
Rewrite the equation as $\frac{6}{x^{2}+3 x+2}+\frac{8}{x^{2}+3 x-4}-1=0$ and let $x^{2}+3 x=y$, then we have:
$\frac{6}{y+2}+\frac{8}{y-4}-1=0 \Leftrightarrow \frac{y(y-16)}{(y+2)(y-4)}=0$, from wh... | x_{1}=0,x_{2}=-3,x_{3,4}=\frac{-3\\sqrt{73}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,218 |
6.148. $3\left(x-\frac{1}{x}\right)+2\left(x^{2}+\frac{1}{x^{2}}\right)=4$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
6.148. $3\left(x-\frac{1}{x}\right)+2\left(x^{2}+\frac{1}{x^{2}}\right)=4$. | ## Solution.
Domain of definition: $x \neq 0$.
Rewrite the equation as $3\left(x-\frac{1}{x}\right)+2\left(x^{2}+\frac{1}{x^{2}}-2\right)=0$ and let $x-\frac{1}{x}=y$. Then we get $2 y^{2}+3 y=0$, from which $y_{1}=0, y_{2}=-\frac{3}{2}$. Therefore, $x-\frac{1}{x}=0$ or $x-\frac{1}{x}=-\frac{3}{2}$.
Solving these eq... | x_{1,2}=\1,x_{3}=-2,x_{4}=0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,219 |
6.149. $\frac{x^{2}+1}{x}+\frac{x}{x^{2}+1}=-2.5$. | Solution.
Domain of definition: $x \neq 0$.
Let $\frac{x^{2}+1}{x}=y$. We get $y+\frac{1}{y}=-2.5 \Leftrightarrow y^{2}+2.5 y+1=0$, from which $y_{1}=-2, y_{2}=-\frac{1}{2}$. Then $\frac{x^{2}+1}{x}=-2$ or $\frac{x^{2}+1}{x}=-\frac{1}{2}$. From this, $x^{2}+2 x+1=0$ or $2 x^{2}+x+2=0$ for $x \neq 0$. Solving these eq... | x_{1,2}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,220 |
6.150. $\frac{u^{2}}{2-u^{2}}+\frac{u}{2-u}=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}u \neq 2, \\ u \neq \pm \sqrt{2} .\end{array}\right.$
From the condition we have $\frac{u^{2}(2-u)+u\left(2-u^{2}\right)-2\left(2-u^{2}\right)(2-u)}{\left(2-u^{2}\right)(2-u)}=0 \Leftrightarrow$ $\Leftrightarrow \frac{2 u^{3}-3 u^{2}-3 u+4}{\left(2-u^{2}\right)(... | u_{1}=1;u_{2,3}=\frac{1\\sqrt{33}}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,221 |
6.151. $\frac{x-m}{x-1}+\frac{x+m}{x+1}=\frac{x-2 m}{x-2}+\frac{x+2 m}{x+2}-\frac{6(m-1)}{5}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq \pm 1, \\ x \neq \pm 2 .\end{array}\right.$
From the condition we have $\frac{(x-m)(x+1)+(x+m)(x-1)}{(x-1)(x+1)}=$
$$
=\frac{(x-2 m)(x+2)+(x+2 m)(x-2)}{(x-2)(x+2)}-\frac{6(m-1)}{5} \Leftrightarrow
$$
$$
\begin{aligned}
& \Leftrightarrow \frac{2 x^{2}... | if\=1,\then\x\\in\R,\except\\1,\\2;\if\\\neq\1,\then\no\roots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,222 |
6.152. $\frac{z+4}{z-1}+\frac{z-4}{z+1}=\frac{z+8}{z-2}+\frac{z-8}{z+2}+6$. | Solution.
Domain of definition: $\left\{\begin{array}{l}z \neq \pm 1, \\ z \neq \pm 2 .\end{array}\right.$
From the condition we get:
$$
\begin{aligned}
& \frac{(z+4)(z+1)+(z-4)(z-1)}{(z-1)(z+1)}=\frac{(z+8)(z+2)+(z-8)(z-2)}{(z-2)(z+2)}+6 \Leftrightarrow \\
& \Leftrightarrow \frac{2 z^{2}+8}{z^{2}-1}=\frac{2 z^{2}+3... | noroots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,223 |
6.153. $(2 x+a)^{5}-(2 x-a)^{5}=242 a^{5}$. | ## Solution.
$$
\begin{gathered}
a^{5}-b^{5}=(a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)=(a-b)\left(\left(a^{4}+b^{4}\right)+\right. \\
\left.+\left(a^{3} b+a b^{3}\right)+a^{2} b^{2}\right)=(a-b)\left(\left((a-b)^{2}+2 a b\right)^{2}-2 a^{2} b^{2}+a b\left(a^{2}+b^{2}\right)+\right. \\
\left.+a^{2} b^{2... | x_{1,2}=\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,224 |
6.154. $\frac{x^{2}+2 x+1}{x^{2}+2 x+2}+\frac{x^{2}+2 x+2}{x^{2}+2 x+3}=\frac{7}{6}$. | ## Solution.
Let $x^{2}+2 x=y$, then the equation in terms of $y$ becomes $\frac{y+1}{y+2}+\frac{y+2}{y+3}-\frac{7}{6}=0 \Leftrightarrow \frac{6(y+1)(y+3)+6(y+2)^{2}-7(y+2)(y+3)}{6(y+2)(y+3)}=0 \Leftrightarrow$
$\Leftrightarrow \frac{5 y^{2}+13 y}{(y+2)(y+3)}=0,5 y^{2}+13 y=0$ for $y \neq-2$ and $y \neq-3 ; y(5 y+13)=... | x_{1}=0,x_{2}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,225 |
6.155. $a x^{4}-x^{3}+a^{2} x-a=0$.
6.155. $a x^{4}-x^{3}+a^{2} x-a=0$. | ## Solution.
We have $\left(a x^{4}-x^{3}\right)+\left(a^{2} x-a\right)=0 \Leftrightarrow x^{3}(a x-1)+a(a x-1)=0 \Leftrightarrow$ $\Leftrightarrow(a x-1)\left(x^{3}+a\right)=0$. Then $a x-1=0$, from which $x_{1}=\frac{1}{a}$, or $x^{3}+a=0$, from which $x_{3}=\sqrt[3]{-a}=-\sqrt[3]{a}$ when $a \neq 0 ; x=0$ when $a=0... | x_{1}=\frac{1}{},x_{2}=-\sqrt[3]{}when\neq0;0when=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,226 |
6.156. $20\left(\frac{x-2}{x+1}\right)^{2}-5\left(\frac{x+2}{x-1}\right)^{2}+48 \frac{x^{2}-4}{x^{2}-1}=0$. | ## Solution.
Domain of definition: $x \neq \pm 1$.
By dividing the equation by $\frac{x^{2}-4}{x^{2}-1} \neq 0$, we have
$$
\begin{aligned}
& 20 \frac{(x-2)^{2}}{(x+1)^{2}} \cdot \frac{x^{2}-1}{x^{2}-4}-5 \frac{(x+2)^{2}}{(x-1)^{2}} \cdot \frac{x^{2}-1}{x^{2}-4}+48=0 \Leftrightarrow \\
& \Leftrightarrow 20 \frac{(x-... | x_{1}=\frac{2}{3},x_{2}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,227 |
6.157. $2(x-1)^{2}-5(x-1)(x-a)+2(x-a)^{2}=0$. | ## Solution.
Dividing the equation by $(x-a)^{2} \neq 0$, we have
$$
2\left(\frac{x-1}{x-a}\right)^{2}-5\left(\frac{x-1}{x-a}\right)+2=0
$$
Let $\frac{x-1}{x-a}=y$.
The equation in terms of $y$ becomes $2 y^{2}-5 y+2=0$, from which $y_{1}=\frac{1}{2}, y_{2}=2$.
Then $\frac{x-1}{x-a}=\frac{1}{2}, x_{1}=2-a$; or $\f... | x_{1}=2-x_{2}=2-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,228 |
6.158. $\sqrt[3]{9-\sqrt{x+1}}+\sqrt[3]{7+\sqrt{x+1}}=4$. | ## Solution.
Domain of definition: $x+1 \geq 0$ or $x \geq-1$.
Let $\sqrt{x+1}=y \geq 0$. The equation in terms of $y$ becomes $\sqrt[3]{9-y}+\sqrt[3]{7+y}=4$. Raising both sides of the equation to the third power, we get
$$
\begin{aligned}
& 9-y+\sqrt[3]{(9-y)^{2}(7+y)}+3 \sqrt[3]{(9-y)(7+y)^{2}}+7+y=64 \Leftrighta... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,229 |
6.159. $\sqrt{x+2}-\sqrt[3]{3 x+2}=0$.
6.159. $\sqrt{x+2}-\sqrt[3]{3 x+2}=0$. | ## Solution.
Domain of definition: $x+2 \geq 0, x \geq-2$.
$$
\begin{aligned}
& \text { From the condition } \sqrt{x+2}=\sqrt[3]{3 x+2} \Rightarrow(x+2)^{3}=(3 x+2)^{2} \\
& x^{3}+6 x^{2}+12 x+8=9 x^{2}+12 x+4 \Leftrightarrow x^{3}-3 x^{2}+4=0 \Leftrightarrow \\
& \Leftrightarrow(x+1)\left(x^{2}-x+1\right)-3(x-1)(x+1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,230 |
6.160. $\sqrt{\frac{20+x}{x}}+\sqrt{\frac{20-x}{x}}=\sqrt{6}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\frac{20+x}{x} \geq 0, \\ \frac{20-x}{x} \geq 0,\end{array} \Leftrightarrow\left\{\begin{array}{l}x(x+20) \geq 0, \\ x(x-20) \geq 0, \\ x \neq 0,0<x \leq 20 .\end{array}\right.\right.$
By squaring both sides of the equation, we have
$$
\begin{aligned}
& \frac{2... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,231 |
6.162. $\sqrt[7]{(a x-b)^{3}}-\sqrt[7]{(b-a x)^{-3}}=\frac{65}{8}(a \neq 0)$. | ## Solution.
Domain of definition: $x \neq \frac{b}{a}$.
## Rewrite the equation as
$$
\sqrt[3]{(a x-b)^{3}}-\frac{1}{\sqrt[7]{(b-a x)^{3}}}=\frac{65}{8} \Leftrightarrow \sqrt[7]{(a x-b)^{3}}+\frac{1}{\sqrt[7]{(a x-b)^{3}}}-\frac{65}{8}=0
$$
Let $\sqrt[7]{(a x-b)^{3}}=y \neq 0$. The equation in terms of $y$ becomes... | x_{1}=\frac{1+128b}{128},x_{2}=\frac{128+b}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,233 |
6.163. $5 \sqrt[15]{x^{22}}+\sqrt[15]{x^{14} \cdot \sqrt{x}}-22 \sqrt[15]{x^{7}}=0$. | ## Solution.
Domain of definition: $x \geq 0$.
From the condition $5 x^{\frac{22}{15}}+x^{\frac{27}{30}}-22 x^{\frac{7}{15}}=0 \Leftrightarrow 5 x^{\frac{44}{30}}+x^{\frac{27}{30}}-22 x^{\frac{14}{30}}=0 \Leftrightarrow$ $\Leftrightarrow x^{\frac{14}{30}}\left(5 x+x^{\frac{1}{2}}-22\right)=0$. Then $x^{\frac{14}{30}}... | x_{1}=0,x_{2}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,234 |
6.164. $\sqrt{x+8+2 \sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4$. | Solution.
Let $\sqrt{x+7}=y \geq 0, x+7=y^{2}, x=y^{2}-7$.
With respect to $y$, the equation takes the form
$$
\begin{aligned}
& \sqrt{y^{2}+2 y+1}+\sqrt{y^{2}-y-6}=4, \sqrt{(y+1)^{2}}+\sqrt{y^{2}-y-6}=4 \\
& |y+1|+\sqrt{y^{2}-y-6}=4
\end{aligned}
$$
Since $y \geq 0$, we have $y+1+\sqrt{y^{2}-y-6}=4, \sqrt{y^{2}-y-... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,235 |
6.165. $\sqrt{\frac{18-7 x-x^{2}}{8-6 x+x^{2}}}+\sqrt{\frac{8-6 x+x^{2}}{18-7 x-x^{2}}}=\frac{13}{6}$. | ## Solution.
Let's write the equation as
$$
\begin{aligned}
& \sqrt{-\frac{x^{2}+7 x-18}{x^{2}-6 x+8}}+\sqrt{-\frac{x^{2}-6 x+8}{x^{2}+7 x-18}}=\frac{13}{6} \Leftrightarrow \\
& \Leftrightarrow \sqrt{-\frac{(x+9)(x-2)}{(x-4)(x-2)}}+\sqrt{-\frac{(x-4)(x-2)}{(x+9)(x-2)}}=\frac{13}{6} \Leftrightarrow \\
& \Leftrightarro... | x_{1}=0,x_{2}=-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,236 |
6.166. $(x+4)(x+1)-3 \sqrt{x^{2}+5 x+2}=6$. | Solution.
Rewrite the equation as $x^{2}+5 x+4-\sqrt{x^{2}+5 x+2}=6 \Leftrightarrow$ $\Leftrightarrow x^{2}+5 x-2-3 \sqrt{x^{2}+5 x+2}=0$. Let $\sqrt{x^{2}+5 x+2}=y$, where $y \geq 0$, $x^{2}+5 x+2=y^{2}, x^{2}+5 x=y^{2}-2$. The equation in terms of $y$ is $y^{2}-3 y-4=0$, from which $y_{1}=-1, y_{2}=4 ; y_{1}=-1<0$ i... | x_{1}=-7,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,237 |
6.167. $\sqrt{x^{2}+x+4}+\sqrt{x^{2}+x+1}=\sqrt{2 x^{2}+2 x+9}$. | Solution.
From the condition $\sqrt{x^{2}+x+4}+\sqrt{x^{2}+x+1}=\sqrt{2\left(x^{2}+x\right)+9}$.
Let $x^{2}+x=y$. The equation in terms of $y$ becomes
$$
\sqrt{y+4}+\sqrt{y+1}=\sqrt{2 y+9}
$$
Squaring both sides of the equation, we get
$y+4+2 \sqrt{(y+4)(y+1)}+y+1=2 y+9 \Leftrightarrow \sqrt{(y+4)(y+1)}=2 \Leftrig... | x_{1}=0,x_{2}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,238 |
6.168. $\sqrt{3 x^{2}-2 x+15}+\sqrt{3 x^{2}-2 x+8}=7$. | ## Solution.
Let $3 x^{2}-2 x=y$. With respect to $y$, the equation takes the form $\sqrt{y+15}+\sqrt{y+8}=7$. Squaring both sides of the equation, we get $y+15+2 \sqrt{(y+15)(y+8)}+y+8=49 \Leftrightarrow \sqrt{(y+15)(y+8)}=13-y$, where $13-y \geq 0, y \leq 13$. From this, $(y+15)(y+8)=(13-y)^{2}, y=1$.
Then $3 x^{2}... | x_{1}=-\frac{1}{3},x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,239 |
6.169. $\sqrt{x}+\frac{2 x+1}{x+2}=2$.
6.169. $\sqrt{x}+\frac{2 x+1}{x+2}=2$. | ## Solution.
Domain of definition: $x \geq 0$.
From the condition, we have $\sqrt{x}=2-\frac{2 x+1}{x+2} \Leftrightarrow \sqrt{x}=\frac{2 x+4-2 x-1}{x+2}, \sqrt{x}=\frac{3}{x+2}$.
Squaring both sides of the equation, we get $x=\frac{3}{x^{2}+4 x+4} \Leftrightarrow$ $\Leftrightarrow \frac{x^{3}+4 x^{2}+4 x-9}{x^{2}+4... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,240 |
6.170. $\frac{\sqrt{x+4}+\sqrt{x-4}}{2}=x+\sqrt{x^{2}-16}-6$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+4 \geq 0, \\ x-4 \geq 0,\end{array} \Leftrightarrow x \geq 4\right.$.
By squaring both sides of the equation, we get
$$
\begin{aligned}
& \frac{x+4+2 \sqrt{(x+4)(x-4)}+x-4}{4}=\left(x+\sqrt{x^{2}-16}\right)^{2}-12\left(x+\sqrt{x^{2}-16}\right)+36 \\
& 2\le... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,241 |
6.171. $\sqrt[3]{x}+\sqrt[3]{x-16}=\sqrt[3]{x-8}$
6.171. $\sqrt[3]{x}+\sqrt[3]{x-16}=\sqrt[3]{x-8}$ | ## Solution.
After raising both sides of the equation to the third power, we have
$$
x+3 \sqrt[3]{x^{2}(x-16)}-3 \sqrt[3]{x(x-16)^{2}}+x-16=x-8 \Leftrightarrow
$$
$$
\Leftrightarrow 3 \sqrt[3]{x(x-16)}(\sqrt[3]{x}+\sqrt[3]{x-16})=8-x
$$
Since by the condition $\sqrt[3]{x}+\sqrt[3]{x-16}=\sqrt[3]{x-8}$, we have
$$
... | x_{1}=8,x_{2,3}=8\\frac{12\sqrt{21}}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,242 |
6.172. $\left(x+\sqrt{x^{2}-1}\right)^{5} \cdot\left(x-\sqrt{x^{2}-1}\right)^{3}=1$. | ## Solution.
Domain of definition: $x^{2}-1 \geq 0 \Leftrightarrow x \in(-\infty ;-1] \cup[1 ; \infty)$.
From the condition we have
$$
\begin{aligned}
& \left(x+\sqrt{x^{2}-1}\right)^{3} \cdot\left(x+\sqrt{x^{2}-1}\right)^{2} \cdot\left(x-\sqrt{x^{2}-1}\right)^{3}=1 \Leftrightarrow \\
& \Leftrightarrow\left(\left(x+... | x_{1}=1,x_{2}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,243 |
6.173. $2 \sqrt{5 \sqrt[4]{x+1}+4}-\sqrt{2 \sqrt[4]{x+1}-1}=\sqrt{20 \sqrt[4]{x+1}+5}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x+1 \geq 0, \\ 2 \sqrt[4]{x+1}-1 \geq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq-1, \\ 2 \sqrt[4]{x+1} \geq 1,\end{array} \Leftrightarrow x \geq-\frac{15}{16}\right.\right.$.
Let $\sqrt[4]{x+1}=y$, where $y \geq 0$. The equation in terms of $y$ be... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,244 |
6.174. $\frac{z}{z+1}-2 \sqrt{\frac{z+1}{z}}=3$. | Solution.
Domain of definition: $\frac{z+1}{z}>0 \Leftrightarrow z \in(-\infty ;-1) \cup(0 ; \infty)$.
Let $\sqrt{\frac{z+1}{z}}=y$, where $y>0$. The equation in terms of $y$ becomes $\frac{1}{y^{2}}-2 y-3=0 \Leftrightarrow 2 y^{3}+3 y^{2}-1=0 \Leftrightarrow 2 y^{3}+2 y^{2}+y^{2}-1=0 \Leftrightarrow$ $\Leftrightarro... | -\frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,245 |
6.175. $\sqrt[3]{x-1}+\sqrt[3]{x-2}-\sqrt[3]{2 x-3}=0$. | ## Solution.
From the condition $\sqrt[3]{x-1}+\sqrt[3]{x-2}=\sqrt[3]{2 x-3}$, raising both sides to the cube, we have
$$
\begin{aligned}
& x-1+3 \sqrt[3]{(x-1)^{2}(x-2)}+3 \sqrt[3]{(x-1)(x-2)^{2}}+x-2=2 x-3 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt[3]{(x-1)(x-2)}(\sqrt[3]{(x-1)}+\sqrt[3]{(x-2)})=0
\end{aligned}
$... | x_{1}=1,x_{2}=2,x_{3}=\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,246 |
6.176. $(\sqrt{x+1}+\sqrt{x})^{3}+(\sqrt{x+1}+\sqrt{x})^{2}=2$.
6.176. $(\sqrt{x+1}+\sqrt{x})^{3}+(\sqrt{x+1}+\sqrt{x})^{2}=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x+1 \geq 0, \\ x \geq 0\end{array} \Leftrightarrow x \geq 0\right.$.
Let $\sqrt{x+1}+\sqrt{x}=y$, where $y \geq 0$. The equation in terms of $y$ becomes
$$
\begin{aligned}
& y^{3}+y^{2}-2=0 \Leftrightarrow y^{3}-1+y^{2}-1=0 \Leftrightarrow \\
& \Leftrightarrow(... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,247 |
6.177. $\sqrt[3]{x+7}-\sqrt{x+3}=0$.
6.177. $\sqrt[3]{x+7}-\sqrt{x+3}=0$. | ## Solution.
Domain of definition: $x+3 \geq 0, x \geq-3$.
From the condition $\sqrt[3]{x+7}=\sqrt{x+3}$ and raising both sides to the sixth power, we get $(x+7)^{2}=(x+3)^{3} \Leftrightarrow x^{2}+14 x+49=x^{3}+9 x^{2}+27 x+27 \Leftrightarrow$
$\Leftrightarrow x^{3}+8 x^{2}+13 x-22=0$. By checking, we find that $x_... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,248 |
6.178. $\frac{\sqrt{(a-x)^{2}}+\sqrt{(a-x)(b-x)}+\sqrt{(b-x)^{2}}}{\sqrt{(a-x)^{2}}-\sqrt{(a-x)(b-x)}+\sqrt{(b-x)^{2}}}=\frac{7}{3}$. | ## Solution.
Domain of definition: $(a-x)(b-x)>0$.
Rewrite the original equation in the following form
$$
\begin{aligned}
& \frac{3 \sqrt{(a-x)^{2}}+3 \sqrt{(a-x)(b-x)}+3 \sqrt{(b-x)^{2}}}{3\left(\sqrt{(a-x)^{2}}-\sqrt{(a-x)(b-x)}+\sqrt{(b-x)^{2}}\right)}+ \\
& +\frac{-7 \sqrt{(a-x)^{2}}+7 \sqrt{(a-x)(b-x)}-7 \sqrt{... | x_{1}=\frac{4-b}{3},x_{2}=\frac{4b-}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,249 |
6.179. $|x|+|x-1|=1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.179. $|x|+|x-1|=1$. | ## Solution.
Let's plot the roots of the absolute values on the number line and consider the equation on the three intervals of the axis. We have three cases:
1) if $x<0$, then $-x-x+1=1$, from which we find $x=0$; there are no solutions, since $x<0$ does not belong to the considered interval;
2) if $0 \leq x<1$, the... | x\in[0;1] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,250 |
6.180. $\left(x^{2}+x+1\right)+\left(x^{2}+2 x+3\right)+\left(x^{2}+3 x+5\right)+\ldots$
$$
\ldots+\left(x^{2}+20 x+39\right)=4500
$$ | ## Solution.
The number of addends is 20. Write the equation in the form
$$
\begin{aligned}
& \left(x^{2}+x^{2}+\ldots+x^{2}\right)+(x+2 x+3 x+\ldots+20 x)+(1+3+5+\ldots+39)=4500 \Leftrightarrow \\
& 20 x^{2}+x(1+2+3+\ldots+20)+(1+3+5+\ldots+39)=4500
\end{aligned}
$$
By the formula for the sum of the terms of an ari... | x_{1}=-\frac{41}{2},x_{2}=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,251 |
6.181. $\sqrt[3]{x+a}+\sqrt[3]{x+a+1}+\sqrt[3]{x+a+2}=0$. | ## Solution.
Rewriting the equation as $\sqrt[3]{x+a}+\sqrt[3]{x+a+1}=-\sqrt[3]{x+a+2}$ and cubing both sides, we get
$$
\begin{aligned}
& x+a+3 \sqrt[3]{(x+a)^{2}(x+a+1)}+3 \sqrt[3]{(x+a)(x+a+1)^{2}}+x+a+1=-(x+a+2) \\
& \sqrt[3]{(x+a)(x+a+1)}(\sqrt[3]{x+a}+\sqrt[3]{x+a+1})=-(x+a+1)
\end{aligned}
$$
Since $\sqrt[3]{... | -(+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,252 |
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