Cho x, y, z đôi một khác nhau và 1/x + 1/y + 1/z = 0. Tính giá trị của biểu thức: A = yz/(x^2 + 2yz) + xz/(y^2 + 2zx) + xy/(z^2 + 2xy)

Đáp án:

$A=1$

Giải thích các bước giải:

$\frac{1}{x}$ $+$ $\frac{1}{y}$ $+$ $\frac{1}{z}$$=0$ $→$$\frac{xy+yz+xz}{xyz}$ $ =0$ $→xy+yz+xz=0$ $→yz=-xy-xz$

$ta có : x² + 2yz = x²+yz-xy-xz=x(x-y)-z(x-y)=(x-y)(x-z)$ $Tương tự: y²+2xz=(y-x)(y-z) ; z²+2xy=(z-x)(z-y)$ $A=$$\frac{yz}{x²+2yz}$ $+$ $\frac{xz}{y²+2xz}$ $+$ $\frac{xy}{z²+xy}$

$=$$\frac{yz}{(x-y)(x-z)}$ $+$ $\frac{xz}{(y-x)(y-z)}$ $+$ $\frac{xy}{(z-x)(z-y)}$

$=$ $\frac{-yz(y-z)}{(x-y)(y-z)(z-x)}$$-$$\frac{xz(z-x)}{(x-y)(y-z)(z-x)}$$-$$\frac{xy(x-y)}{(x-y)(y-z)(z-x)}$

$=$ $\frac{-yz(y-z)-xz(z-x)-xy(x-y)}{(x-y)(y-z)(z-x)}$

$=$ $\frac{-yz(y-z)+xz(y-z+x-y)-xy(x-y)}{(x-y)(y-z)(z-x)}$

$=$$\frac{-yz(y-z)+xz(y-z)+xz(x-y)-xy(x-y)}{(x-y)(y-z)(z-x)}$ $=$$\frac{(y-z)(xz-yz)+(x-y)(xz-xy)}{(x-y)(y-z)(z-x)}$

$=$$\frac{z(x-y)(y-z)-x(x-y)(y-z)}{(x-y)(y-z)(z-x)}$ $=$$\frac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x}$ $=1$