Question

\(3x^2+y^2+z^2+2x-2y+2xy+3=0\),x=?,y=?,z=?

Answer 1

Lời giải:

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

\(\Leftrightarrow (x^2+y^2+1+2xy-2y-2x)+2(x^2+2x+1)+z^2=0\)

\(\Leftrightarrow (x+y-1)^2+2(x+1)^2+z^2=0\)

Vì \(\left\{\begin{matrix} (x+y-1)^2\geq 0\\ (x+1)^2\geq 0\\ z^2\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{R}\)

Do đó: \((x+y-1)^2+2(x+1)^2+z^2\geq 0\)

Dấu bằng xảy ra khi \(\left\{\begin{matrix} (x+y-1)^2=0\\ (x+1)^2=0\\ z^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-1\\ y=2\\ z=0\end{matrix}\right.\)