Question

Rút gọn \(\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\) , biết rằng \(x+y+z=0\)

Answer 1

Có:\(x+y+z=0\)

\(\Rightarrow\left(x+y+z\right) ^2=0\)

\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)

\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\)

Có:

\(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\)

\(=y^2-2yz+z^2+z^2-2xz+z^2+x^2-2xy+y^{^2}\)

\(=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)

\(=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2\)

\(=3\left(x^2+y^2+z^2\right)\)

\(\Rightarrow\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)

\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\)

\(=\dfrac{1}{3}\)