Bài 105428

Question

Cho $xy,yz,zx \neq -1$.Chứng minh rằng:
$\frac{x-y}{1+xy}+\frac{y-z}{1+yz}+\frac{z-x}{1+zx}=\frac{x-y}{1+xy}.\frac{y-z}{1+yz}. \frac{z-x}{1+zx}$

score 0 · Answer 1

Đặt: $x=\tan a, y=\tan b, z=\tan c (-\frac{\pi}{2}<a,b,c<\frac{\pi}{2})$
Khi đó $\begin{cases}\frac{x-y}{1+xy}=\frac{\tan a-\tan b}{1+\tan a\tan b}=\tan (a-b) \\\frac{y-z}{1+yz} =\frac{\tan b-\tan c}{1+\tan b\tan c}=\tan (b-c)\\\frac{z-x}{1+zx}=\frac{\tan c-\tan a}{1+\tan c\tan a}=\tan (c-a) \end{cases}$
Hơn nữa: $(a-b)+(b-c)=-(c-a)$
$\Rightarrow \tan [(a-b)+(b-c)]=-\tan(c-a)$
$\Rightarrow \frac{\tan (a-b)+\tan(b-c)}{1-\tan (a-b).\tan(b-c)}=- \tan(c-a)$
$\Rightarrow \tan (a-b)+\tan(b-c)+\tan(c-a)=\tan (a-b).\tan(b-c).\tan(c-a)$
$\Rightarrow \frac{x-y}{1+xy}+\frac{y-z}{1+yz}+\frac{z-x}{1+zx}=\frac{x-y}{1+xy}.\frac{y-z}{1+yz}.\frac{z-x}{1+zx}$

score 0 · Answer 2

Đặt: $x=\tan a, y=\tan b, z=\tan c (-\frac{\pi}{2}<a,b,c<\frac{\pi}{2})$
Khi đó $\begin{cases}\frac{x-y}{1+xy}=\frac{\tan a-\tan b}{1+\tan a\tan b}=\tan (a-b) \\\frac{y-z}{1+yz} =\frac{\tan b-\tan c}{1+\tan b\tan c}=\tan (b-c)\\\frac{z-x}{1+zx}=\frac{\tan c-\tan a}{1+\tan c\tan a}=\tan (c-a) \end{cases}$
Hơn nữa: $(a-b)+(b-c)=-(c-a)$
$\Rightarrow \tan [(a-b)+(b-c)]=-\tan(c-a)$
$\Rightarrow \frac{\tan (a-b)+\tan(b-c)}{1-\tan (a-b).\tan(b-c)}=- \tan(c-a)$
$\Rightarrow \tan (a-b)+\tan(b-c)+\tan(c-a)=\tan (a-b).\tan(b-c).\tan(c-a)$
$\Rightarrow \frac{x-y}{1+xy}+\frac{y-z}{1+yz}+\frac{z-x}{1+zx}=\frac{x-y}{1+xy}.\frac{y-z}{1+yz}.\frac{z-x}{1+zx}$

Bài 105428

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Bài 105428

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