Lời giải:
Ta có:
$2x^3+x^2+4x+2=(2x+1)(x^2+2)=\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}\geq 0$
$\Rightarrow 2x+1\geq 0$
$\Rightarrow x\geq \frac{-1}{2}$
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\sqrt{x^2-\frac{1}{4}+\sqrt{(x+\frac{1}{2})^2}}=\sqrt{x^2-\frac{1}{4}+|x+\frac{1}{2}|}\)
\(=\sqrt{x^2-\frac{1}{4}+x+\frac{1}{2}}\) (do $x\geq \frac{-1}{2}$)
\(=\sqrt{x^2+x+\frac{1}{4}}=\sqrt{(x+\frac{1}{2})^2}=|x+\frac{1}{2}|=x+\frac{1}{2}\)
PT đã cho trở thành:
$x+\frac{1}{2}=(2x+1)(x^2+2)$
$\Leftrightarrow 2x+1=2(2x+1)(x^2+2)$
$\Leftrightarrow (2x+1)(2x^2+3)=0$
$\Leftrightarrow 2x+1=0$ (do $2x^2+3>0$)
$\Leftrightarrow x=-\frac{1}{2}$ (tm)
