Question

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

Answer 1

Ta có: \(\dfrac{1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(y-z\right)\left(z-x\right)}+\dfrac{1}{\left(z-x\right)\left(x-y\right)}\)

\(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-z\right)\left(x-y\right)}\)

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

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

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