Câu hỏi của nguyễn phùng phước

Question

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

nhanh nha...sáng thứ 3 mình thi rồi

Quìn · Accepted Answer

Vì $\left(x-y^2+z\right)^2\ge0$

$\left(y-2\right)^2\ge0$

$\left(z-3\right)^2\ge0$

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

$\Rightarrow$ $\left(x-y^2+z\right)^2=0;\text{ }\left(y-2\right)^2=0;\text{ }\left(z-3\right)^2=0$

+$\text{ }\left(y-2\right)^2=0$

$\Rightarrow\text{ }y-2=0$

$y=0+2$

$y=2$

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

$\Rightarrow z-3=0$

$z=0+3$

$z=3$

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

$\Rightarrow x-y^2+z=0$

$x-2^2+3=0$

$x-4=0-3$

$x-4=-3$

$x=-3+4$

$x=1$

Vậy: $x=1;\text{ }y=2;\text{ }z=3$Câu trả lời: Vì $\left(x-y^2+z\right)^2\ge0$