4:
1/x+1/y+1/z=0
=>(xy+yz+xz)/xyz=0
=>xy+yz+xz+0
=>yz=-xy-xz
x^2+2yz=x^2+yz-xy-xz
=(x-y)(x-z)
Tương tự, ta sẽ có: y^2+2xz=(y-x)(y-z)
z^2+2xy=(z-x)(z-y)
\(A=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-x\right)\cdot\left(y-z\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=1\)