a) Ta có: \(A=\frac{1}{2x^2+2x+5}\)
\(=\frac{1}{2\left(x^2+x+\frac{5}{2}\right)}=\frac{1}{2\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{9}{4}\right)}=\frac{1}{2\left(x+\frac{1}{2}\right)^2+\frac{9}{2}}\)
Ta có: \(2\left(x+\frac{1}{2}\right)^2+\frac{9}{2}\ge\frac{9}{2}\forall x\)
\(\Rightarrow\frac{1}{2\left(x+\frac{1}{2}\right)^2+\frac{9}{2}}\le\frac{2}{9}\forall x\)
Dấu '=' xảy ra khi
\(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy: GTLN của biểu thức \(A=\frac{1}{2x^2+2x+5}\) là \(\frac{2}{9}\) khi \(x=-\frac{1}{2}\)
