Đáp án:
$\begin{array}{l}
6)\\
a)A = \sqrt {\dfrac{{2x - 3}}{{x - 1}}} \\
Dkxd:\dfrac{{2x - 3}}{{x - 1}} \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x - 3 \ge 0\\
x - 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - 3 \le 0\\
x - 1 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x < 1
\end{array} \right.\\
B = \dfrac{{\sqrt {2x - 3} }}{{\sqrt {x - 1} }}\\
Dkxd:\left\{ \begin{array}{l}
2x - 3 \ge 0\\
x - 1 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x > 1
\end{array} \right. \Leftrightarrow x \ge \dfrac{3}{2}\\
b)Dkxd:\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x < 1
\end{array} \right.\\
x \ge \dfrac{3}{2}
\end{array} \right. \Leftrightarrow x \ge \dfrac{3}{2}\\
A = B\\
\Leftrightarrow \sqrt {\dfrac{{2x - 3}}{{x - 1}}} = \dfrac{{\sqrt {2x - 3} }}{{\sqrt {x - 1} }}\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - 3 \ge 0\\
x - 1 \ge 0
\end{array} \right. \Leftrightarrow x \ge \dfrac{3}{2}\\
Vậy\,x \ge \dfrac{3}{2}\\
7)\\
a)\sqrt {\dfrac{{289}}{{225}}} = \dfrac{{17}}{{15}}\\
b)\sqrt {2\dfrac{{14}}{{25}}} = \sqrt {\dfrac{{64}}{{25}}} = \dfrac{8}{5}\\
c)\sqrt {\dfrac{{0,25}}{9}} = \dfrac{{0,5}}{3} = \dfrac{1}{{2.3}} = \dfrac{1}{6}\\
d)\sqrt {1\dfrac{9}{{16}}.5\dfrac{4}{9}.0,01} \\
= \sqrt {\dfrac{{25}}{{16}}.\dfrac{{49}}{9}.0,01} \\
= \dfrac{5}{4}.\dfrac{7}{3}.0,1\\
= \dfrac{7}{{24}}\\
e)\sqrt {\dfrac{{{{149}^2} - {{76}^2}}}{{{{457}^2} - {{384}^2}}}} \\
= \sqrt {\dfrac{{\left( {149 - 76} \right)\left( {149 + 76} \right)}}{{\left( {457 + 384} \right)\left( {457 - 384} \right)}}} \\
= \sqrt {\dfrac{{73.225}}{{841.73}}} \\
= \dfrac{{15}}{{29}}
\end{array}$
