Question

Cho x,y,z đôi một khác nhau và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

Tính \(A=\dfrac{yz}{x^2+2xy}+\dfrac{zx}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)

Answer 1

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Leftrightarrow\dfrac{yz+xz+xy}{xyz}=0\)

\(\Leftrightarrow xy+xz+yz=0\)

~ ~ ~

\(x^2+2yz\)

\(=x^2+yz-xy-xz\)

\(=\left(x-y\right)\left(x-z\right)\)

Tương tự, ta có: \(y^2+2xz=\left(y-x\right)\left(y-z\right)\) và \(z^2+2xy=\left(z-x\right)\left(z-y\right)\)

\(A=\dfrac{yz}{\left(x-z\right)\left(x-y\right)}+\dfrac{xz}{\left(y-x\right)\left(y-z\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\)

\(A=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}\)

\(=\dfrac{\left(x-z\right)\left(x-y\right)\left(y-z\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}\)

= 1