Question

tìm x,y,z biết :(x-y²+z)²+(y-2)²+(z+3)²=0

Answer 1

\(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)

Vì \(\left(x-y^2+z\right)^2\ge0;\left(y-2\right)^2\ge0;\left(x+3\right)^2\ge0\)nên \(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2\ge0\)

Mà \(\left(x-y^2+z\right)+\left(y-2\right)^2+\left(x+3\right)^2=0\)nên \(\hept{\begin{cases}x-y^2+z=0\\y-2=0\\z+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=7\\y=2\\z=-3\end{cases}}}\)