Đáp án:

$S = \{0;2\}$

Giải thích các bước giải:

$\begin{array}{l}x^2 + x + 5 = 2\sqrt{4x + 1} + \sqrt{8x+9}\qquad \left(ĐK:x \geq -\dfrac14\right)\\ \Leftrightarrow x(x+1) = 2(\sqrt{4x+1} -1) + \sqrt{8x+9} -3\\ \Leftrightarrow x(x+1) = 2\cdot\dfrac{(\sqrt{4x+1} -1)(\sqrt{4x+1} +1)}{\sqrt{4x+1}+1} +\dfrac{(\sqrt{8x+9} -3)(\sqrt{8x+9} +3)}{\sqrt{8x+9} +3}\\ \Leftrightarrow x(x+1) = \dfrac{8x}{\sqrt{4x+1}+1}+\dfrac{8x}{\sqrt{8x+9} +3}\\\\ \Leftrightarrow x\left(x + 1 -\dfrac{8}{\sqrt{4x+1}+1} -\dfrac{8}{\sqrt{8x+9} +3}\right)=0\\ \Leftrightarrow \left[\begin{array}{l}x = 0\quad (nhận)\\x + 1 =\dfrac{8}{\sqrt{4x+1}+1} +\dfrac{8}{\sqrt{8x+9} +3}\quad (*)\end{array}\right.\\ Giải \,\,(*):\\ (*) \Leftrightarrow x - 2 = \dfrac{8}{\sqrt{4x+1} + 1} - 2 + \dfrac{8}{\sqrt{8x+9}+3}-1\\ \Leftrightarrow x - 2 = \dfrac{6 - 2\sqrt{4x+1}}{\sqrt{4x+1} + 1} + \dfrac{5 - \sqrt{8x+9}}{\sqrt{8x+9}+3}\\ \Leftrightarrow x - 2 =\dfrac{2(3 - \sqrt{4x+1})(3 + \sqrt{4x+1})}{(\sqrt{4x+1} + 1)(3 +\sqrt{4x+1})} + \dfrac{(5 - \sqrt{8x+9})(5 + \sqrt{8x+9})}{(\sqrt{8x+9}+3)(5 + \sqrt{8x+9})}\\ \Leftrightarrow x - 2 = \dfrac{2(8 - 4x)}{(\sqrt{4x+1} + 1)(3 +\sqrt{4x+1})} + \dfrac{16 - 8x}{(\sqrt{8x+9}+3)(5 + \sqrt{8x+9})}\\ \Leftrightarrow x - 2 =\dfrac{8(2-x)}{(\sqrt{4x+1} + 1)(3 +\sqrt{4x+1})} + \dfrac{8(2-x)}{(\sqrt{8x+9}+3)(5 + \sqrt{8x+9})}\\ \Leftrightarrow (x-2)\left[1 + \dfrac{8}{(\sqrt{4x+1} + 1)(3 +\sqrt{4x+1})} +\dfrac{8}{(\sqrt{8x+9}+3)(5 + \sqrt{8x+9})} \right]=0\\ \Leftrightarrow \left[\begin{array}{l} x = 2\quad (nhận)\\1 + \dfrac{8}{(\sqrt{4x+1} + 1)(3 +\sqrt{4x+1})} +\dfrac{8}{(\sqrt{8x+9}+3)(5 + \sqrt{8x+9})}=0\quad (**)\end{array}\right.\\ \text{Dễ dàng nhận thấy}\\ 1 + \dfrac{8}{(\sqrt{4x+1} + 1)(3 +\sqrt{4x+1})} +\dfrac{8}{(\sqrt{8x+9}+3)(5 + \sqrt{8x+9})} >0,\quad \forall x \geq -\dfrac14\\ \text{Do đó $(**)$ vô nghiệm}\\ \text{Vậy phương trình có tập nghiệm}\,\,S=\{0;2\} \end{array}$

Đáp án:

Giải thích các bước giải:

 ĐKXĐ: $x \geq -\dfrac{1}{4}$

$⇔(x^2-2x)+(x+3-\sqrt{8x+9})+2(x+1-\sqrt{4x+1})=0$

$⇔(x^2-2x)+\dfrac{x^2-2x}{x+3+\sqrt{8x+9}}+\dfrac{2(x^2-2x)}{x+1+\sqrt{4x+1}}=0$

$⇔(x^2-2x)\left ( 1+\dfrac{1}{x+3+\sqrt{8x+9}}+\dfrac{2}{x+1+\sqrt{4x+1}} \right )=0$

$⇔x^2-2x=0$ (ngoặc phía sau luôn dương)

$⇔x=0$ hoặc $x=2$