Đáp án:
$\begin{array}{l}
1)a)Dkxd:x \ge \dfrac{1}{2}\\
\sqrt {{x^2}} = 2x - 1\\
\Leftrightarrow {x^2} = 4{x^2} - 4x + 1\\
\Leftrightarrow 3{x^2} - 4x + 1 = 0\\
\Leftrightarrow \left( {3x - 1} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow x = 1\left( {tm} \right)\left( {do:x \ge \dfrac{1}{2}} \right)\\
Vậy\,x = 1\\
b)Dkxd:x \ge - 2\\
\sqrt {{x^2} - 4x + 4} = x + 2\\
\Leftrightarrow {x^2} - 4x + 4 = {x^2} + 4x + 4\\
\Leftrightarrow 8x = 0\\
\Leftrightarrow x = 0\left( {tm} \right)\\
Vậy\,x = 0\\
c)\sqrt {{{\left( {x - 2} \right)}^2}} = 4\\
\Leftrightarrow {\left( {x - 2} \right)^2} = 16\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 4\\
x - 2 = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = - 2
\end{array} \right.\\
Vậy\,x = - 2;x = 6\\
d)\sqrt {{{\left( {3 + x} \right)}^2}} = \sqrt 9 \\
\Leftrightarrow {\left( {3 + x} \right)^2} = 9\\
\Leftrightarrow \left[ \begin{array}{l}
3 + x = 3\\
3 + x = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 6
\end{array} \right.\\
Vậy\,x = 0;x = - 6\\
2)a)\sqrt {{3^2}} - \sqrt {{{\left( { - 7} \right)}^2}} + \sqrt {{{\left( { - 1} \right)}^2}} \\
= 3 - 7 + 1\\
= - 3\\
b)2\sqrt {{{\left( { - 2} \right)}^2}} + 3\sqrt {{{\left( { - 5} \right)}^2}} + \sqrt {{3^2}} \\
= 2.2 + 3.5 + 3\\
= 22\\
c)\sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {2 + \sqrt 2 } \right)}^2}} \\
= 2 - \sqrt 2 + 2 + \sqrt 2 \\
= 4\\
d)\sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} \\
= 3 - \sqrt 2 - \left( {\sqrt 2 - 1} \right)\\
= 4 - 2\sqrt 2 \\
e)\sqrt {3 - 2\sqrt 2 } + \sqrt {3 + 2\sqrt 2 } \\
= \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} \\
= \sqrt 2 - 1 + \sqrt 2 + 1\\
= 2\sqrt 2 \\
f)\sqrt {4 - 2\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \sqrt 3 - 1 - \left( {\sqrt 3 + 1} \right)\\
= - 2
\end{array}$
