Đáp án:
Trong câu hỏi trước chị đã trả lời câu c -> g nên giờ chị trả lời ý h và i.
$\begin{array}{l}
h)Dkxd:x \ge 3\\
\sqrt {x - 2 - 2\sqrt {x - 3} } - \sqrt {x + 1 - 4\sqrt {x - 3} } = 2x - 5\\
\Rightarrow \sqrt {x - 3 - 2\sqrt {x - 3} + 1} \\
- \sqrt {x - 3 - 4\sqrt {x - 3} + 4} = 2x - 5\\
\Rightarrow \sqrt {{{\left( {\sqrt {x - 3} - 1} \right)}^2}} - \sqrt {{{\left( {\sqrt {x - 3} - 2} \right)}^2}} = 2x - 5\\
\Rightarrow \left| {\sqrt {x - 3} - 1} \right| - \left| {\sqrt {x - 3} - 2} \right| = 2x - 5\\
+ Khi:\sqrt {x - 3} - 2 \ge 0\\
\Rightarrow \sqrt {x - 3} \ge 2\\
\Rightarrow x \ge 7\\
\Rightarrow \sqrt {x - 3} - 1 - \sqrt {x - 3} + 2 = 2x - 5\\
\Rightarrow 2x = 6\\
\Rightarrow x = 3\left( {ktm} \right)\\
+ Khi:1 \le \sqrt {x - 3} < 2\\
\Rightarrow 4 \le x < 7\\
\Rightarrow \sqrt {x - 3} - 1 + \sqrt {x - 3} - 2 = 2x - 5\\
\Rightarrow 2x - 2\sqrt {x - 3} - 2 = 0\\
\Rightarrow x - \sqrt {x - 3} - 1 = 0\\
\Rightarrow \sqrt {x - 3} = x - 1\\
\Rightarrow x - 3 = {x^2} - 2x + 1\\
\Rightarrow {x^2} - 3x + 4 = 0\left( {\text{vô}\,\text{nghiệm}} \right)\\
+ Khi:x < 4\\
\Rightarrow 1 - \sqrt {x - 3} + \sqrt {x - 3} - 2 = 2x - 5\\
\Rightarrow x = 2\left( {tm} \right)\\
\text{Vậy}\,x = 2\\
i)Dkxd:x \ge - \dfrac{1}{4}\left( 1 \right)\\
x + \sqrt {x + \dfrac{1}{2} + \sqrt {x + \dfrac{1}{4}} } = \dfrac{1}{4}\\
\Rightarrow \sqrt {x + \dfrac{1}{4} + 2.\sqrt {x + \dfrac{1}{4}} .\dfrac{1}{2} + \dfrac{1}{4}} = \dfrac{1}{4} - x\left( {dk:x \le \dfrac{1}{4}} \right)\\
\Rightarrow \sqrt {{{\left( {\sqrt {x + \dfrac{1}{4}} + \dfrac{1}{2}} \right)}^2}} = \dfrac{1}{4} - x\\
\Rightarrow \sqrt {x + \dfrac{1}{4}} + \dfrac{1}{2} = \dfrac{1}{4} - x\\
\Rightarrow \sqrt {x + \dfrac{1}{4}} = - x - \dfrac{1}{4}\\
\Rightarrow Dkxd: - x - \dfrac{1}{4} \ge 0\\
\Rightarrow x \le - \dfrac{1}{4}\left( 2 \right)\\
\text{Từ}\,\left( 1 \right);\left( 2 \right) \Rightarrow x = - \dfrac{1}{4}
\end{array}$
