Đáp án:
$S =\{-9+\sqrt{41};0\}$
Giải thích các bước giải:
$\quad (x+5)\sqrt{2x^2 +1} = x^2 + x + 5$
$\to (x+5)(\sqrt{2x^2 +1} -1) = x^2$
$\to (x+5)\dfrac{(\sqrt{2x^2 +1} -1)(\sqrt{2x^2 +1} +1)}{\sqrt{2x^2 +1} +1}=x^2$
$\to \dfrac{2x^2(x+5)}{\sqrt{2x^2 +1} +1}=x^2$
$\to x^2\left[\dfrac{2(x+5)}{\sqrt{2x^2 +1} +1} - 1\right]=0$
$\to \left[\begin{array}{l}x = 0\\\dfrac{2(x+5)}{\sqrt{2x^2 +1} +1} = 1\qquad (*)\end{array}\right.$
$(*)\Leftrightarrow 2(x+5) = \sqrt{2x^2 +1} +1$
$\Leftrightarrow 2x + 9 = \sqrt{2x^2 +1}$
$\Leftrightarrow \begin{cases}2x + 9 > 0\\(2x + 9)^2 = 2x^2 + 1\end{cases}$
$\Leftrightarrow \begin{cases}x > -\dfrac92\\x^2 + 18x + 40= 0\end{cases}$
$\Leftrightarrow \begin{cases}x > -\dfrac92\\\left[\begin{array}{l}x = -9 -\sqrt{41}\quad (loại)\\x = -9 + \sqrt{41} \quad (nhận)\end{array}\right.\end{cases}$
Vậy $S =\{-9+\sqrt{41};0\}$
