Ta có: \(2x^2+10x-1\)
\(=2\left(x^2+5x-\frac{1}{2}\right)\)
\(=2\left(x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{16}-\frac{33}{16}\right)\)
\(=2\left[\left(x+\frac{5}{2}\right)^2-\frac{33}{16}\right]\)
\(=2\left(x+\frac{5}{2}\right)^2-\frac{33}{8}\)
Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x+\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x+\frac{5}{2}\right)^2-\frac{33}{8}\le\frac{33}{8}\forall x\)
Dấu '=' xảy ra khi
\(2\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\)
\(\Leftrightarrow x=\frac{-5}{2}\)
vậy: GTLN của đa thức \(2x^2+10x-1\) là \(\frac{33}{8}\) khi x=\(\frac{-5}{2}\)