Do \(\left(x+2y\right)^2\ge0;\left(y-1\right)^2\ge0;\left(x-z\right)^2\ge0\forall x;y;z\)
Mà theo đề bài: \(\left(x+2y\right)^2+\left(y-1\right)^2+\left(x-z\right)^2=0\)
=> \(\begin{cases}\left(x+2y\right)^2=0\\\left(y-1\right)^2=0\\x-z=0\end{cases}\)=> \(\begin{cases}x+2y=0\\y-1=0\\x=z\end{cases}\)=> \(\begin{cases}x=-2y\\y=1\\x=z\end{cases}\)
=> x = z = -2; y = 1
Ta có:
x + 2y + 3z = -2 + 2.2 + 3.(-2)
= -2 + 4 + (-6)
= 2 + (-6)
= -4
\(\left(x+2y\right)^2+\left(y-1\right)^2+\left(x-z\right)^2=0\)
\(x^2+4xy+4y^2+y^2-2y+1+x^2-2xz+z^2=0\)
\(2x^2+5y^2+z^2+4xy-2y-2xz=0\)
Đến đây thì mk chịu
