Question

cho x+y+z=-3, x^2+y^2+z^2=xy+yz+zx tính x^2+y^3+z^4

Answer 1

\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\) \(\Leftrightarrow x=y=z\)

Mà \(x+y+z=-3\Rightarrow x=y=z=-1\)

\(\Rightarrow x^2+y^3+z^4=\left(-1\right)^2+\left(-1\right)^3+\left(-1\right)^4=1\)