Question

Với x + y + z = 0. Rút gọn:

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

Answer 1

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

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

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

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

\(=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}\)

\(=1\)

Vậy \(x+y+z=0\)thì \(A=1\)

Answer 2

Cho mình hỏi tại sao x + y = -z \(\Rightarrow\)x-y = z