Ta có:` x+y+z=7`
`⇔z=-x-y+7`
`⇔xy+z-6=xy-x-y+7-6`
`⇔xy+z-6=xy-x-y+1`
`⇔xy+z-6=x(y-1)-(y-1)`
`⇔xy+z-6=(y-1)(x-1)`
Tương tự ta có:
`yz+x-6=(y-1)(z-1)`
`xz+y-6=(x-1)(z-1)`
Ta lại có:
`(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)`
`⇔7^2=23+2(xy+xz+yz)`
$⇔xy+xz+yz=\dfrac{49-23}{2}=13$
$H=\dfrac{1}{xy+z-6}+\dfrac{1}{yz+x-6}+\dfrac{1}{xz+y-6}$
$⇔H=\dfrac{1}{(y-1)(x-1)}+\dfrac{1}{(y-1)(z-1)}+\dfrac{1}{(x-1)(z-1)}$
$⇔H=\dfrac{z-1+x-1+y-1}{(y-1)(x-1)(z-1)}$
$⇔H=\dfrac{z+x+y-3}{(xy-y-x+1)(z-1)}$
$⇔H=\dfrac{7-3}{xyz-xy+y-yz-xz+x+z-1}$
$⇔H=\dfrac{4}{(x+y+z)-(xy+xz+yz)+xyz-1}$
$⇔H=\dfrac{4}{7-(xy+xz+yz)+3-1}$
$⇔H=\dfrac{4}{9-(xy+xz+yz)}$
$⇔H=\dfrac{4}{9-13}$
$⇔H=\dfrac{4}{-4}=-1$
Vậy `H=-1`
