\(P\left(x\right)=\dfrac{4x^4+16x^3+56x^2+80x+356}{x^2+2x+5}\\ P\left(x\right)=\dfrac{4x^2\left(x^2+2x+5\right)+8x\left(x^2+2x+5\right)+20\left(x^2+2x+5\right)+256}{x^2+2x+5}\\ P\left(x\right)=4\left(x^2+2x+5\right)+\dfrac{256}{x^2+2x+5}\\ \ge2\sqrt{\dfrac{4\left(x^2+2x+5\right)\cdot256}{x^2+2x+5}}=2\sqrt{1024}=64\left(BĐTcosi\right)\)
Dấu \("="\Leftrightarrow4\left(x^2+2x+5\right)=\dfrac{256}{x^2+2x+5}\)
\(\Leftrightarrow x^2+2x+5=8\Leftrightarrow x^2+2x-3=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
P(x)=\(\dfrac{\text{(4x^2+8x^3+20x^2)+(8x^3+16x^2+40x)+(20x^2+40x+100)+256}}{x^2+2x+5}\)
=(4x^2+8x+20x) +\(\dfrac{256}{x^2+2x+5}\)
áp dụng BĐT Cosi a+b≥\(2\sqrt{ab}\)
=>P(x)≥64
Dấu = xảy ra khi x=-1 hoặc x=3
