Đáp án:
`S = {1;5}`
Giải thích các bước giải:
\(\sqrt{x+2\sqrt{x-1}} + \sqrt{x-2\sqrt{x-1}}\\(DK:x \ge 1) \\\Leftrightarrow 2\sqrt{x+2\sqrt{x-1}} +2\sqrt{x-2\sqrt{x-1}} = x+3\\\Leftrightarrow 4(x+2\sqrt{x-1})+8\sqrt{(x+2\sqrt{x-1})(x-2\sqrt{x-1})} +4(x-2\sqrt{x-1})=x^2 + 6x +9\\\Leftrightarrow 4x + 8\sqrt{x-1} +8\sqrt{x^2 -4(x-1)}+4x - 8\sqrt{x-1}=x^2 + 6x +9\\\Leftrightarrow 8\sqrt{x^2 - 4x +4} = x^2 - 2x +9\\\Leftrightarrow 8\sqrt{(x-2)^2} = x^2 - 2x +9\\\Leftrightarrow 8|x-2|=x^2 - 2x +9\\\Leftrightarrow \left[ \begin{array}{l}8(x-2)-x^2 + 2x = 9, \quad x-2\ge 0\\\\8[-(x-2)]-x^2 +2x =9, \quad x-2\le 0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}10x - 16 - x^2 =9,\quad x\ge 2\\\\-6x + 16 - x^2 =9,\quad x < 2\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}-(x-5)^2=0, \quad x\ge 2\\\\(x+7)(x-1)=0, \quad x<2\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=5\\\\\left[ \begin{array}{l}x=1\\\\x=-7\quad(\text{ktm})\end{array} \right.\end{array} \right.\\\text{Vậy}\quad x=1\quad \text{hoặc} \quad x =5\)
