\(A=\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+1+\dfrac{y^2-xz}{\left(y+z\right)\left(y+x\right)}+1+\dfrac{z^2-xy}{\left(z+x\right)\left(z+y\right)}+1-3\)
Xét \(\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+1=\dfrac{x^2-yz+x^2+xz+xy+yz}{\left(x+y\right)\left(x+z\right)}\)
\(=\dfrac{x^2+xy+x^2+xz}{\left(x+y\right)\left(x+z\right)}=\dfrac{x\left(x+y\right)+x\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}=\dfrac{x}{x+y}+\dfrac{x}{x+z}\)
Tương tự: \(\left\{{}\begin{matrix}\dfrac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+1=\dfrac{y}{y+z}+\dfrac{y}{y+x}\\\dfrac{z^2-xy}{\left(z+y\right)\left(z+x\right)}+1=\dfrac{z}{z+y}+\dfrac{z}{z+x}\end{matrix}\right.\)
Cộng vế với vế ta được:
\(A=\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{y+x}+\dfrac{y}{y+z}+\dfrac{z}{z+x}+\dfrac{z}{z+y}-3\)
\(A=\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}-3=1+1+1-3=0\)