Đáp án:
$x = 1\,hoac\,x = 5 + 4\sqrt 2 $
Giải thích các bước giải:
$\begin{array}{l}
\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } = \frac{{x + 3}}{2}\\
dkxd:\left\{ \begin{array}{l}
x \ge 1\\
x - 2\sqrt {x - 1} \ge 0\left( {luon\,dung} \right)
\end{array} \right. \Rightarrow x \ge 1\\
pt \Leftrightarrow \sqrt {x - 1 + 2\sqrt {x - 1} + 1} + \sqrt {x - 1 - 2\sqrt {x - 1} + 1} = \frac{{x + 3}}{2}\\
\Rightarrow \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} = \frac{{x + 3}}{2}\\
\Rightarrow \sqrt {x - 1} + 1 + \left| {\sqrt {x - 1} - 1} \right| = \frac{{x + 3}}{2}\\
+ TH1:\,\sqrt {x - 1} - 1 \ge 0 \Rightarrow x \ge 2\\
thi\,pt \Rightarrow \sqrt {x - 1} + 1 + \sqrt {x - 1} - 1 = \frac{{x + 3}}{2}\\
\Rightarrow 2\sqrt {x - 1} = \frac{{x + 3}}{2}\\
\Rightarrow 16\left( {x + 1} \right) = {x^2} + 6x + 9\\
\Rightarrow {x^2} - 10x - 7 = 0\\
\Rightarrow x = 5 + 4\sqrt 2 \left( {do\,x \ge 2} \right)\\
+ TH2:\,\sqrt {x - 1} - 1 < 0 \Rightarrow 1 \le x < 2\\
thi\,pt \Rightarrow \sqrt {x - 1} + 1 - \sqrt {x - 1} + 1 = \frac{{x + 3}}{2}\\
\Rightarrow 2 = \frac{{x + 3}}{2} \Rightarrow x = 1\left( {tmdk} \right)\\
vay\,x = 1\,hoac\,x = 5 + 4\sqrt 2
\end{array}$
