Question

Cho \(x+y+z=0\). Tính \(A=\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)

Answer 1

Ta có :\(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2=0\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2zx=0\)

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

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

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

\(A=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2-\left(x^2+y^2+z^2+2xy+2yz+2zx\right)}\)

\(A=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)

Answer 2

x+y+z=0

x=y=z=0=> A ko xac dinh

kl

gt cua A tren may tinh duoc