Đáp án:

$\begin{array}{l}
3)DKxd:x \ne 1\\
\dfrac{9}{{x - 1}} = \dfrac{{x + 1}}{7}\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 1} \right) = 9.7\\
 \Leftrightarrow {x^2} - 1 = 63\\
 \Leftrightarrow {x^2} = 64\\
 \Leftrightarrow x = 8;x =  - 8\left( {tmdk} \right)\\
Vậy\,x = 8;x =  - 8\\
4)Dkxd:x \ne  - 2;x \ne  - 3\\
\dfrac{{x - 1}}{{x + 2}} = \dfrac{{x - 2}}{{x + 3}}\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 3} \right) = \left( {x - 2} \right)\left( {x + 2} \right)\\
 \Leftrightarrow {x^2} + 2x - 3 = {x^2} - 4\\
 \Leftrightarrow 2x =  - 1\\
 \Leftrightarrow x =  - \dfrac{1}{2}\left( {tmdk} \right)\\
Vậy\,x =  - \dfrac{1}{2}\\
5)Dkxd:\left\{ \begin{array}{l}
x \ne  - \dfrac{7}{5}\\
x \ne  - \dfrac{1}{5}
\end{array} \right.\\
\dfrac{{3x + 2}}{{5x + 7}} = \dfrac{{3x - 1}}{{5x + 1}}\\
 \Leftrightarrow \left( {3x + 2} \right)\left( {5x + 1} \right) = \left( {3x - 1} \right)\left( {5x + 7} \right)\\
 \Leftrightarrow 15{x^2} + 3x + 10x + 2 = 15{x^2} + 21x - 5x - 7\\
 \Leftrightarrow 13x + 2 = 16x - 7\\
 \Leftrightarrow 3x = 9\\
 \Leftrightarrow x = 3\left( {tmdk} \right)\\
Vậy\,x = 3\\
6)Dkxd:x \ne 1\\
\left( {x - 1} \right):6\dfrac{1}{4} = \dfrac{1}{9}:\left( {x - 1} \right)\\
 \Leftrightarrow \left( {x - 1} \right):\dfrac{{25}}{4} = \dfrac{1}{9}:\left( {x - 1} \right)\\
 \Leftrightarrow \left( {x - 1} \right).\left( {x - 1} \right) = \dfrac{{25}}{4}.\dfrac{1}{9}\\
 \Leftrightarrow {\left( {x - 1} \right)^2} = \dfrac{{25}}{{36}}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = \dfrac{5}{6}\\
x - 1 =  - \dfrac{5}{6}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{6}\\
x = \dfrac{1}{6}
\end{array} \right.\\
Vậy\,x = \dfrac{{11}}{6};x = \dfrac{1}{6}\\
7))0,75:\left( {3 - 2x} \right) = \left( {2x - 3} \right):\dfrac{{ - 1}}{3}\\
 \Leftrightarrow 0,75.\dfrac{{ - 1}}{3} = \left( {3 - 2x} \right)\left( {2x - 3} \right)\\
 \Leftrightarrow  - \dfrac{1}{4} =  - {\left( {2x - 3} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = \dfrac{1}{2}\\
2x - 3 =  - \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{7}{2}\\
2x = \dfrac{5}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{4}\\
x = \dfrac{5}{4}
\end{array} \right.\\
Vậy\,x = \dfrac{7}{4};x = \dfrac{5}{4}
\end{array}$