Question

\( \dfrac{1}{(x-y)(y-z)}+\dfrac{1}{(y-z)(z-x)}+\dfrac{1}{(z-x)(x-y)}\)

Giải giúp ạ, cần gấp

Answer 1

\(\dfrac{1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(y-z\right)\left(z-x\right)}+\dfrac{1}{\left(z-x\right)\left(x-y\right)}\)(đk: \(x\ne y\ne z\))

\(=\dfrac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\dfrac{0}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

Answer 2

Chắc đề là tính ha!

\(=\dfrac{x+y+y-z+x-y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ =\dfrac{0}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ =0\\ Vậy.A=0\)