Question

[Nguyễn Hải Linh]

a) Tìm GTNN của biểu thức: P=2x²+5y²+4xy+8x-4y+15

Answer 1

[Nguyễn Thanh Hằng]

\(P=2x^2+5y^2+4xy+8x-4y+15\)

\(=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-5\)

\(=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-5\)

Ta có :

\(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) \(\Leftrightarrow P\ge-5\)

Dấu "=" xảy ra khi :

\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)

Vậy \(P_{Min}=-5\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)