\(A=\frac{\left(4x^4+16x^3+16x^2\right)+\left(40x^2+80x\right)+356}{x^2+2x+5}=\frac{4.\left(x^2+2x\right)^2+40\left(x^2+2x\right)+356}{x^2+2x+5}\)
\(=\frac{4\left[\left(x^2+2x\right)^2+10\left(x^2+2x\right)+25\right]+256}{x^2+2x+5}\)\(=\frac{4\left(x^2+2x+5\right)^2+4^4}{x^2+2x+5}=4\left[\left(x^2+2x+5\right)+\frac{4^3}{x^2+2x+5}\right]\)
Áp dụng Côsi:
\(A\ge4.2\sqrt{\left(x^2+2x+5\right).\frac{4^3}{x^2+2x+5}}=64\)
Dấu "=" xảy ra khi \(x^2+2x+5=\frac{4^3}{x^2+2x+5}\Leftrightarrow\left(x^2+2x+5\right)^2=64\Leftrightarrow x^2+2x+5=8\)(do x2+2x+5 > 0)
\(\Leftrightarrow x^2+2x-3=0\Leftrightarrow x=1\text{ hoặc }x=-3\)
Vậy GTNN của A là 64.
