\(-\dfrac{4\sqrt{x}+3}{x+1}=\dfrac{-\left(4\sqrt{x}+3\right)}{x+1}=-4+\dfrac{4x-4\sqrt{x}+1}{x+1}=-4+\dfrac{\left(2\sqrt{x}\right)^2-2.2.\sqrt{x}+1}{x+1}=-4+\dfrac{\left(2\sqrt{x}-1\right)^2}{x+1}\)
\(với:x>0\Rightarrow x+1>1>0\)
\(A=4x+\dfrac{1}{4x}-\dfrac{4\sqrt{x}+3}{x+1}+2016=4x+\dfrac{1}{4x}-4+\dfrac{\left(2\sqrt{x}-1\right)^2}{x+1}+2016\ge2\sqrt{4x.\dfrac{1}{4x}}-4+2016=2-4+2016=2014\)
\(min_A=2014\Leftrightarrow4x=\dfrac{1}{4x}\Leftrightarrow x=\dfrac{1}{4}\)
\(A=4x+\dfrac{1}{4x}+\dfrac{-4\sqrt{x}-3}{x+1}+2016\)
\(=4x+\dfrac{1}{4x}+\dfrac{\left(2\sqrt{x}-1\right)^2-4\left(x+1\right)}{x+1}+2016\)
\(=4x+\dfrac{1}{4x}+\dfrac{\left(2\sqrt{x}-1\right)^2}{x+1}+2012\)
Có \(4x+\dfrac{1}{4x}\ge2\sqrt{4x.\dfrac{1}{4x}}=2\)
\(\dfrac{\left(2\sqrt{x}-1\right)^2}{x+1}\ge0\)
=> \(A=4x+\dfrac{1}{4x}+\dfrac{\left(2\sqrt{x}-1\right)^2}{x+1}+2012\ge2014\)
"=" khi \(\left\{{}\begin{matrix}4x=\dfrac{1}{4x}\\2\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\)
