Question

Giải bất phương trình: \(4x^2-\sqrt{2x^3+2x^2+x+1}>6x+4\)

Answer 1

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow4x^2-6x-4-\sqrt{\left(2x^2+1\right)\left(x+1\right)}>0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow2a^2-6b^2-ab>0\)

\(\Leftrightarrow\left(2a+3b\right)\left(a-2b\right)>0\)

\(\Leftrightarrow a-2b>0\)

\(\Leftrightarrow a>2b\Leftrightarrow\sqrt{2x^2+1}>2\sqrt{x+1}\)

\(\Leftrightarrow2x^2+1>4x+4\)

\(\Leftrightarrow2x^2-4x-3>0\Rightarrow\left[{}\begin{matrix}-1\le x< \frac{2-\sqrt{10}}{2}\\x>\frac{2+\sqrt{10}}{2}\end{matrix}\right.\)