1/a/ \(13x-2x^2=-2\left(x^2-2.\frac{13}{4}x+\frac{169}{16}\right)+\frac{169}{8}=-2\left(x-\frac{13}{4}\right)^2+\frac{169}{8}\le\frac{169}{9}\)
b/ \(-3x^2-8x=-3\left(x^2+2.\frac{4}{3}x+\frac{16}{9}\right)+\frac{16}{3}=-3\left(x+\frac{4}{3}\right)^2+\frac{16}{3}\le\frac{16}{3}\)
Câu 2:
a/ \(x^2+2xy+2y^2+4x+20\)
\(=2\left(\frac{x^2}{4}+xy+y^2\right)+\frac{1}{2}\left(x^2+8x+16\right)+12\)
\(=2\left(\frac{x}{2}+y\right)^2+\frac{1}{2}\left(x+4\right)^2+12\ge12\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
b/ \(5x^2-2x+y^2-2y-4xy+8\)
\(=\left(4x^2+y^2+1-4xy+4x-2y\right)+\left(x^2-6x+9\right)-2\)
\(=\left(2x-y+1\right)^2+\left(x-3\right)^2-2\ge-2\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)