Đáp án:
$\begin{array}{l}
5d)1 = 2 - 1 = \sqrt 4 - 1 > \sqrt 3 - 1\\
\Leftrightarrow 1 > \sqrt 3 - 1\\
6d)\\
Dkxd:\dfrac{{2 - 3x}}{2} \ge 0\\
\Leftrightarrow 2 - 3x \ge 0\\
\Leftrightarrow 3x \le 2\\
\Leftrightarrow x \le \dfrac{2}{3}\\
Vậy\,x \le \dfrac{2}{3}\\
7a)Dkxd:x \ge - 1\\
\Leftrightarrow \sqrt {{x^2} + 1} = x + 1\\
\Leftrightarrow {x^2} + 1 = {x^2} + 2x + 1\\
\Leftrightarrow 2x = 0\\
\Leftrightarrow x = 0\left( {tmdk} \right)\\
Vậy\,x = 0\\
b)Dkxd:x \ge 0\\
\Leftrightarrow \sqrt x < 2\\
\Leftrightarrow x < 4\\
Vậy\,0 \le x < 4\\
c)Dkxd:2x + 1 \ge 0 \Leftrightarrow x \ge - \dfrac{1}{2}\\
\Leftrightarrow \sqrt {9{x^2}} = 2x + 1\\
\Leftrightarrow 9{x^2} = 4{x^2} + 4x + 1\\
\Leftrightarrow 5{x^2} - 4x - 1 = 0\\
\Leftrightarrow \left( {5x + 1} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow x = 1\left( {do:x \ge \dfrac{{ - 1}}{2}} \right)\\
Vậy\,x = 1\\
d)Dkxd:x \ge \dfrac{1}{3}\\
\sqrt {{x^2} + 6x + 9} = 3x - 1\\
\Leftrightarrow \sqrt {{{\left( {x + 3} \right)}^2}} = 3x - 1\\
\Leftrightarrow x + 3 = 3x - 1\\
\Leftrightarrow 2x = 4\\
\Leftrightarrow x = 2\left( {tmdk} \right)\\
Vậy\,x = 2\\
8)\\
a)\sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} = 1 + \sqrt 2 \\
b)\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} = 2 - \sqrt 3 \\
c)\sqrt {{{\left( {4 - \sqrt {17} } \right)}^2}} = \sqrt {17} - 4\\
d)\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} = 2 + \sqrt 3 \\
9)\\
a)\sqrt {3 - 2\sqrt 2 } = \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} = \sqrt 2 - 1\\
b)\sqrt {7 + 2\sqrt {10} } = \sqrt {{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}} = \sqrt 5 + \sqrt 2 \\
c)\sqrt {15 - 6\sqrt 6 } = \sqrt {{{\left( {3 - \sqrt 6 } \right)}^2}} = 3 - \sqrt 6 \\
d)\sqrt {17 + 12\sqrt 2 } = \sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} = 3 + 2\sqrt 2
\end{array}$
