Đáp án:
$\frac{1+\sqrt{13}}{2}$
Giải thích các bước giải:
ĐKXĐ: \(x\ge-\frac{3}{2}\)
\(pt\Leftrightarrow\left(2x+5\right)\sqrt{2x+3}=8x^3+4x\)
\(\Leftrightarrow\left(2x+3\right)\sqrt{2x+3}+2\sqrt{2x+3}=8x^3+4x\left(1\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\\2x=b\end{matrix}\right.\left(a\ge0;b\ge-3\right)\)
\(\left(1\right)\Leftrightarrow a^3+2a=b^3+2b\)
\(\Leftrightarrow a^3-b^3+2a-2b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+2\right)=0\)
\(\Leftrightarrow a-b=0\left(\text{Vì }a^2+ab+b^2+2>0\right)\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{2x+3}=2x\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-2x-3=0\\x\ge0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{1+\sqrt{13}}{2}\left(TMĐK\right)\)
