Question

Answer 1

Câu 39:

\(E=x^2+4y^2+z^2-4x-6z+4y+16\)

\(=x^2-4x+4+4y^2+4y+1+z^2-6z+9+2\)

\(=\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-3\right)^2+2\ge2\forall x,y,z\)

Dấu '=' xảy ra khi \(\begin{cases}x-2=0\\ 2y+1=0\\ z-3=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=-\frac12\\ z=3\end{cases}\)