Giải thích các bước giải:
Ta có:
$\dfrac1{x-y}\cdot \dfrac1{y-z}+\dfrac1{y-z}\cdot \dfrac1{z-x}+\dfrac1{z-x}\cdot \dfrac1{x-y}=\dfrac{z-x+x-y+y-z}{(x-y)(y-z)(z-x)}=0$
Khi đó:
$\dfrac1{(x-y)^2}+\dfrac{1}{(y-z)^2}+\dfrac{1}{(z-x)^2}$
$=\dfrac1{(x-y)^2}+\dfrac{1}{(y-z)^2}+\dfrac{1}{(z-x)^2}+2(\dfrac1{x-y}\cdot \dfrac1{y-z}+\dfrac1{y-z}\cdot \dfrac1{z-x}+\dfrac1{z-x}\cdot \dfrac1{x-y})$
$=(\dfrac1{x-y}+\dfrac1{y-z}+\dfrac1{z-x})^2$
Là bình phương của $2$ số hữu tỉ