Đáp án:
$S=1$
Giải thích các bước giải:
Ta có:
`1/x+1/y+1/z=0`
`->xyz(1/x+1/y+1/z)=0`
`->xy+yz+zx=0`
`->`$\begin{cases}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{cases}$
Ta có:
`x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x(x-y)-z(x-y)=(x-y)(x-z)`
Tương tự, ta được:
$\begin{cases}y^2+2zx=(y-x)(y-z)\\z^2+2xy=(z-x)(z-y)\end{cases}$
Thay vào $S$, ta có:
$S=\dfrac{yz-3}{(x-y)(x-z)}+\dfrac{zx-3}{(y-x)(y-z)}+\dfrac{xy-3}{(z-x)(z-y)}$
$=\dfrac{3-yz}{(x-y)(z-x)}+\dfrac{3-zx}{(x-y)(y-z)}+\dfrac{3-xy}{(z-x)(y-z)}$
$=\dfrac{(3-yz)(y-z)}{(x-y)(y-z)(z-x)}+\dfrac{(3-zx)(z-x)}{(x-y)(y-z)(z-x)}+\dfrac{(3-xy)(x-y)}{(x-y)(y-z)(z-x)}$
$=\dfrac{3y-3z-y^2z+yz^2+3z-3x-z^2x+zx^2+3x-3y-x^2y+xy^2}{(x-y)(y-z)(z-x)}$
$=\dfrac{-y^2z+yz^2-z^2x+zx^2-x^2y+xy^2}{(x-y)(y-z)(z-x)}$
$=\dfrac{(zx^2-y^2z)-(z^2x-yz^2)-(x^2y-xy^2)}{(x-y)(y-z)(z-x)}$
$=\dfrac{z(x-y)(x+y)-z^2(x-y)-xy(x-y)}{(x-y)(y-z)(z-x)}$
$=\dfrac{(x-y)[z(x+y)-z^2-xy]}{(x-y)(y-z)(z-x)}$
$=\dfrac{(x-y)(xz+yz-z^2-xy)}{(x-y)(y-z)(z-x)}$
$=\dfrac{(x-y)[-z(z-x)+y(z-x)]}{(x-y)(y-z)(z-x)}$
$=\dfrac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=1$
