ĐKXĐ: \(x>0\)
\(5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=t>0\)
\(\Rightarrow t^2=x+\dfrac{1}{4x}+1\Rightarrow x+\dfrac{1}{4x}=t^2-1\)
BPT trở thành:
\(5t< 2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2>0\)
\(\Leftrightarrow\left[{}\begin{matrix}t>2\\t< \dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\dfrac{1}{2\sqrt{x}}>2\\\sqrt{x}+\dfrac{1}{2\sqrt{x}}< \dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1>0\\2x-\sqrt{x}+1< 0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}>\dfrac{2+\sqrt{2}}{2}\\\sqrt{x}< \dfrac{2-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{3+2\sqrt{2}}{2}\\0< x< \dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\)
