Question

[Thị Châu Nguyễn Phạm]

vòng 16

Answer 1

[Hoàng Thị Ngọc Anh]

Ta có: \(\left[{}\begin{matrix}\left(x-3y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(x+z\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-3y\right)^2+\left(y-1\right)^2+\left(x+z\right)^2\ge0\)

Dấu \("="\) xảy ra khi: \(\left[{}\begin{matrix}\left(x-3y\right)^2=0\\\left(y-1\right)^2=0\\\left(x+z\right)^2=0\end{matrix}\right.\)'\(\Rightarrow\left[{}\begin{matrix}x=3\\y=1\\z=-3\end{matrix}\right.\)

Khi đó: \(A=3.3+2.1+\left(-3\right)=8\)

Vậy \(A=8.\)

Answer 2

[Hoàng Thị Ngọc Anh]

=8 nhé!