`1.`

`(x-y)(x-z)`

`=x(x-z)-y(x-z)`

`=x^2-zx-xy+yz`

`=x^2-zx-xy+yz+xy+yz+zx`

`=x^2+2yz` (đpcm)

`2.`

Tương tự câu `1`, suy ra:

`y^2+2zx=(y-z)(y-x)`

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

Suy ra:

`1/{x^2+2yz}+1/{y^2+2zx}+1/{z^2+2xy}`

`=1/{(x-y)(x-z)}+1/{(y-x)(y-z)}+1/{(z-x)(z-y)}`

`={y-z}/{(x-y)(x-z)(y-z)}+{z-x}/{(x-y)(x-z)(y-z)}+{x-y}/{(x-y)(x-z)(y-z)}`

`={y-z+z-x+x-y}/{(x-y)(x-z)(y-z)}`

`=0` (đpcm)