Đáp án:
$\begin{array}{l}
1)a)Dkxd:3 + x \ge 0\\
\Leftrightarrow x \ge - 3\\
Vậy\,x \ge - 3\\
b)\dfrac{{ - 3}}{{x - 1}} \ge 0\\
\Leftrightarrow x - 1 < 0\\
\Leftrightarrow x < 1\\
Vậy\,x < 1\\
2)a)\sqrt {50} .\sqrt 2 - \sqrt {32} :\sqrt 2 \\
= \sqrt {100} - \sqrt {16} \\
= 10 - 4\\
= 6\\
b)\sqrt {{{\left( {2 - \sqrt 5 } \right)}^2}} - \sqrt {{{\left( {4 + \sqrt 5 } \right)}^2}} \\
= \sqrt 5 - 2 - \left( {4 + \sqrt 5 } \right)\\
= \sqrt 5 - 2 - 4 - \sqrt 5 \\
= - 6\\
3)a)Dkxd:x \ge 1\\
\sqrt {4x - 4} = 6\\
\Leftrightarrow 4x - 4 = 36\\
\Leftrightarrow 4x = 40\\
\Leftrightarrow x = 10\left( {tmdk} \right)\\
Vậy\,x = 10\\
b)Dkxd:x \ge \dfrac{3}{2}\\
\dfrac{{\sqrt {2x - 3} }}{{\sqrt {x - 1} }} = 2\\
\Leftrightarrow \sqrt {2x - 3} = 2\sqrt {x - 1} \\
\Leftrightarrow 2x - 3 = 4\left( {x - 1} \right)\\
\Leftrightarrow 4x - 2x = - 3 + 4\\
\Leftrightarrow 2x = 1\\
\Leftrightarrow x = \dfrac{1}{2}\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
4)a)x > 0\\
A = \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} - \dfrac{6}{x}\sqrt {\dfrac{{{x^2}}}{4}} \\
= \sqrt x + 1 - \dfrac{6}{x}.\dfrac{x}{2}\\
= \sqrt x + 1 - 3\\
= \sqrt x - 2\\
b)x = 7 - 4\sqrt 3 \left( {tmdk} \right)\\
= 4 - 2.2.\sqrt 3 + 3\\
= {\left( {2 - \sqrt 3 } \right)^2}\\
\Leftrightarrow \sqrt x = 2 - \sqrt 3 \\
\Leftrightarrow A = \sqrt x - 2 = 2 - \sqrt 3 - 2 = - \sqrt 3
\end{array}$
