cho x y z từng đôi 1 khác nhau va1/x+1/y+1/x=0 tính B=(yz/x^2+2yz)+(xz/y^2+2xz)+(xy/z^2+2xy)

Đáp án: $B=1$

Giải thích các bước giải:

 $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\rightarrow xy+yz+zx=0$

$\rightarrow x^2+2yz=x^2+yz-xy-xz=x(x-y)-z(x-y)=(x-y)(x-z)$

Tương tự ta có :

$y^2+2xz=(y-z)(y-z)$

$z^2+2xy=(z-x)(z-y)$

$\rightarrow B=\dfrac{yz}{(x-y)(x-z)}+\dfrac{xz}{(y-x)(y-x)}+\dfrac{xy}{(z-x)(z-y)}$

$\rightarrow -B=\dfrac{yz(y-z)}{(x-y)(y-z)(z-x)}+\dfrac{xz(z-x)}{(x-y)(y-z)(z-x)}+\dfrac{xy(x-y)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{yz(y-z)+xz(z-x)+xy(x-y)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{y^2z-yz^2+xz^2-x^2z+xy(x-y)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{y^2z-x^2z-(yz^2-xz^2)+xy(x-y)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{z(y^2-x^2)-z^2(y-x)+xy(x-y)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{z(y-x)(y+x)-z^2(y-x)+xy(x-y)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{(x-y)(-z(y+x)+z^2)+xy(x-y)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{(x-y)(-z(y+x)+z^2+xy)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{(x-y)(-zy-zx+z^2+xy)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{(x-y)(z(z-x)-y(z-x)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=\dfrac{(x-y)(z-y)(z-x)}{(x-y)(y-z)(z-x)}$

$\rightarrow -B=-1$

$\rightarrow B=1$