Đáp án:
$\begin{array}{l}
1)c)Dkxd:\dfrac{{x - 3}}{{x + 2}} \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x < - 2
\end{array} \right.\\
Vậy\,x \ge 3\,hoặc\,x \le - 2\\
d)Dkxd:\dfrac{{{x^2} + 3}}{{1 - 6x + 9{x^2}}} \ge 0\\
\Leftrightarrow 1 - 6x + 9{x^2} > 0\\
\Leftrightarrow {\left( {3x - 1} \right)^2} > 0\\
\Leftrightarrow 3x - 1\# 0\\
\Leftrightarrow x\# \dfrac{1}{3}\\
Vậy\,x\# \dfrac{1}{3}\\
B3)\\
a)3 = \sqrt 9 < \sqrt {12} = 2\sqrt 3 \\
Vậy\,3 < 2\sqrt 3 \\
b)6 + 2\sqrt 2 = 6 + \sqrt 8 \\
9 = 6 + 3 = 6 + \sqrt 9 > 6 + \sqrt 8 \\
\Leftrightarrow 6 + 2\sqrt 2 < 9\\
c)9 + 4\sqrt 5 = {\left( {2 + \sqrt 5 } \right)^2} > {\left( {2 + \sqrt 4 } \right)^2} = 16\\
\Leftrightarrow 9 + 4\sqrt 5 > 16\\
d)\sqrt 7 + 2 = \sqrt {{{\left( {\sqrt 7 + 2} \right)}^2}} = \sqrt {11 + 2.2\sqrt 7 } \\
= \sqrt {11 + 2.\sqrt {28} } \\
\sqrt 5 + \sqrt 6 = \sqrt {11 + 2\sqrt {30} } > \sqrt {11 + 2.\sqrt {28} } \\
\Leftrightarrow \sqrt 7 + 2 < \sqrt 5 + \sqrt 6 \\
C4)\\
a)\sqrt {{x^2}} = 1\\
\Leftrightarrow {x^2} = 1\\
\Leftrightarrow x = 1;x = - 1\\
Vậy\,x = 1;x = - 1\\
b)Dkxd:x \ge \dfrac{1}{2}\\
\sqrt {2x - 1} = 3\\
\Leftrightarrow 2x - 1 = 9\\
\Leftrightarrow 2x = 10\\
\Leftrightarrow x = 5\\
Vậy\,x = 5\\
c)\sqrt {{x^2} - 2x + 1} = 2\\
\Leftrightarrow {x^2} - 2x + 1 = 4\\
\Leftrightarrow {x^2} - 2x - 3 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow x = 3;x = - 1\\
Vậy\,x = 3;x = - 1\\
d)Dkxd:x \ge - 1\\
\sqrt {{x^2} - 6x + 9} = x + 1\\
\Leftrightarrow {x^2} - 6x + 9 = {x^2} + 2x + 1\\
\Leftrightarrow 8x = 8\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1\\
B5)\\
a)3\sqrt 2 - \sqrt {6 - 4\sqrt 2 } \\
= 3\sqrt 2 - \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} \\
= 3\sqrt 2 - 2 + \sqrt 2 \\
= 4\sqrt 2 - 2\\
b)\sqrt {2.\left( {4 - \sqrt 7 } \right)} - \sqrt 7 \\
= \sqrt {8 - 2\sqrt 7 } - \sqrt 7 \\
= \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} - \sqrt 7 \\
= \sqrt 7 - 1 - \sqrt 7 \\
= - 1\\
c)\sqrt {23 + 8\sqrt 7 } - \sqrt 7 \\
= \sqrt {16 + 2.4.\sqrt 7 + 7} - \sqrt 7 \\
= \sqrt {{{\left( {4 + \sqrt 7 } \right)}^2}} - \sqrt 7 \\
= 4 + \sqrt 7 - \sqrt 7 \\
= 4\\
d)\sqrt {16{x^2}} + 5x\\
= - 4x + 5x\left( {do:x < 0} \right)\\
= x
\end{array}$
