Đáp án:
$\begin{array}{l}
3)\sqrt {{{\left( {2 - \sqrt 5 } \right)}^2}} .\sqrt {{{\left( {\sqrt 2 + \sqrt 3 } \right)}^2}} \\
= \left| {2 - \sqrt 5 } \right|.\left| {\sqrt 2 + \sqrt 3 } \right|\\
= \left( {\sqrt 5 - 2} \right).\left( {\sqrt 2 + \sqrt 3 } \right)\\
= \sqrt {10} - \sqrt {15} - 2\sqrt 2 - 2\sqrt 3 \\
4)\sqrt {{{\left( {1 - 2\sqrt 3 } \right)}^2}} \\
= \left| {1 - 2\sqrt 3 } \right|\\
= 2\sqrt 3 - 1\\
5)\sqrt {{{\left( {\sqrt 5 - 2\sqrt 3 } \right)}^2}} \\
= \left| {\sqrt 5 - 2\sqrt 3 } \right|\\
= 2\sqrt 3 - \sqrt 5 \\
6)\sqrt {{{\left( {\sqrt 7 - 3\sqrt 2 } \right)}^2}} \\
= \left| {\sqrt 7 - 3\sqrt 2 } \right|\\
= 3\sqrt 2 - \sqrt 7 \\
B3)\\
a)Dkxd:3x - 2 \ge 0\\
\Leftrightarrow 3x \ge 2\\
\Leftrightarrow x \ge \dfrac{2}{3}\\
Vậy\,x \ge \dfrac{2}{3}\\
b)\dfrac{{2x}}{y}\sqrt {\dfrac{y}{x}} \\
Dkxd:\left\{ \begin{array}{l}
y\# 0\\
\dfrac{y}{x} \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y\# 0\\
x.y > 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x > 0;y > 0\\
x < 0;y < 0
\end{array} \right.\\
Vậy\,x < 0;y < 0\,hoặc\,x > 0;y > 0\\
B4)\\
a)\sqrt {{x^2} - 2x + 1} = 7\\
\Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2}} = 7\\
\Leftrightarrow \left| {x - 1} \right| = 7\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 7\\
x - 1 = - 7
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = - 6
\end{array} \right.\\
Vậy\,x = - 6;x = 8\\
b)\sqrt {9{x^2} - 6x + 1} = 6\\
\Leftrightarrow \sqrt {{{\left( {3x - 1} \right)}^2}} = 6\\
\Leftrightarrow \left| {3x - 1} \right| = 6\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 1 = 6\\
3x - 1 = - 6
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x = - \dfrac{5}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 5}}{3};x = \dfrac{7}{3}
\end{array}$
