Đáp án: $x=1$
Giải thích các bước giải:
ĐKXĐ: $x\ge -\dfrac13$
Ta có:
$x^2-2(x+1)\sqrt{3x+1}=2\sqrt{2x^2+5x+2}-8x-5$
$\to (x^2+2x+1)-2(x+1)\sqrt{3x+1}+3x+1=2\sqrt{(2x+1)(x+2)}-3x-3$
$\to (x+1)^2-2(x+1)\sqrt{3x+1}+3x+1=-(2x+1-2\sqrt{(2x+1)(x+2)}+x+2)$
$\to (x+1-\sqrt{3x+1})^2=-(\sqrt{2x+1}-\sqrt{x+2})^2$
$\to (x+1-\sqrt{3x+1})^2+(\sqrt{2x+1}-\sqrt{x+2})^2=0$
Mà $(x+1-\sqrt{3x+1})^2+(\sqrt{2x+1}-\sqrt{x+2})^2\ge 0$
$\to$Dấu $=$ xảy ra khi :
$\begin{cases}x+1-\sqrt{3x+1}=0\\ \\\sqrt{2x+1}-\sqrt{x+2}=0\to\sqrt{2x+1}=\sqrt{x+2}\to x=1\end{cases}$
