Bài 105430

Question

Cho $\begin{cases}x+y+z=xyz \\ x,y,z \neq \pm \frac{\sqrt{3}}{3} \end{cases}$. Chứng minh rằng:
$\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2}.\frac{3y-y^3}{1-3y^2}.\frac{3z-z^3}{1-3z^2} (1)$

score 0 · Answer 1

Đặt $ x=\tan \alpha,    y=\tan \beta, z=\tan \gamma     (\alpha,\beta,\gamma \neq \frac{\pi}{2}, k\in Z)$
Từ giả thiết: $x+y+z=xyz$
$\Leftrightarrow \tan \alpha+\tan \beta+\tan \gamma=\tan \alpha.\tan \beta.\tan \gamma$
$\Leftrightarrow \tan \alpha(1-\tan \beta\tan \gamma)=-(\tan \beta+\tan \gamma)$
*   Nếu $ 1-\tan \beta\tan \gamma=0$ thì $yz=1$ $\Rightarrow y+z=0$
$\Rightarrow \begin{cases}VT (1)=\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2} \\ VP (1)= \frac{3x-x^3}{1-3x^2}.\frac{3y-y^3}{1-3y^2}.\frac{\frac{3}{y}-\frac{1}{y^3}}{1-\frac{3}{y^2}}\\           =\frac{3x-x^3}{1-3x^2}.\frac{3y-y^3}{1-3y^2}.\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2}\end{cases}$
$\Rightarrow   VT (1)=VP (1)$
* Nếu $ 1-\tan \beta\tan \gamma \neq 0$ thì:
$\tan \alpha=-\frac{\tan \beta+\tan \gamma}{1-\tan \beta\tan \gamma}               \Leftrightarrow \tan (-\alpha)=\tan (\beta+\gamma)$
$\Leftrightarrow \alpha+\beta+\gamma=k\pi, k\in Z          \Leftrightarrow 3\alpha+3\beta+3\gamma=3k\pi$
$ \Leftrightarrow \tan (-3\alpha)=\tan (3\beta+3\gamma) \Leftrightarrow -\tan 3\alpha=\frac{\tan 3\beta+\tan 3\gamma}{1-\tan 3\beta\tan 3\gamma} $
$\Leftrightarrow   \tan 3\alpha+\tan 3\beta+\tan 3\gamma   = \tan 3\alpha.\tan 3\beta.\tan 3\gamma$
Hơn nữa $ \tan 3\alpha=\frac{3\tan \alpha- \tan^3 \alpha}{1-3\tan^2 \alpha}\Rightarrow$ thay ngược trở lại ta có ĐPCM

Bài 105430

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Bài 105430

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