Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
7,\\
a,\\
\dfrac{{2x}}{3}:\dfrac{5}{{6{x^2}}} = \dfrac{{2x}}{3}.\dfrac{{6{x^2}}}{5} = \dfrac{{4{x^3}}}{5}\\
b,\\
\dfrac{{25{x^3}{y^5}}}{3}:15x{y^2} = \dfrac{{25{x^3}{y^5}}}{3}.\dfrac{1}{{15x{y^2}}} = \dfrac{{5{x^2}{y^3}}}{9}\\
c,\\
\dfrac{{{x^2} - {y^2}}}{{6{x^2}y}}:\dfrac{{x + y}}{{3xy}} = \dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{6{x^2}y}}.\dfrac{{3xy}}{{x + y}} = \dfrac{{x - y}}{{2x}}\\
d,\\
\dfrac{{{a^2} + ab}}{{b - a}}:\dfrac{{a + b}}{{2{a^2} - 2{b^2}}} = \dfrac{{a.\left( {a + b} \right)}}{{b - a}}:\dfrac{{a + b}}{{2.\left( {{a^2} - {b^2}} \right)}}\\
 = \dfrac{{a.\left( {a + b} \right)}}{{b - a}}:\dfrac{{a + b}}{{2.\left( {a - b} \right)\left( {a + b} \right)}}\\
 = \dfrac{{a.\left( {a + b} \right)}}{{b - a}}:\dfrac{1}{{2\left( {a - b} \right)}}\\
 = \dfrac{{a.\left( {a + b} \right)}}{{b - a}}.2.\left( {a - b} \right)\\
 =  - 2a\left( {a + b} \right)\\
8,\\
a,\\
\left( {\dfrac{1}{{{x^2} + x}} - \dfrac{{2 - x}}{{x + 1}}} \right):\left( {\dfrac{1}{x} + x - 2} \right)\\
 = \left( {\dfrac{1}{{x\left( {x + 1} \right)}} + \dfrac{{x - 2}}{{x + 1}}} \right):\dfrac{{1 + {x^2} - 2x}}{x}\\
 = \dfrac{{1 + \left( {x - 2} \right).x}}{{x\left( {x + 1} \right)}}.\dfrac{x}{{{x^2} - 2x + 1}}\\
 = \dfrac{{{x^2} - 2x + 1}}{{x\left( {x + 1} \right)}}.\dfrac{x}{{{x^2} - 2x + 1}}\\
 = \dfrac{1}{{x + 1}}\\
b,\\
\left( {\dfrac{{3x}}{{1 - 3x}} + \dfrac{{2x}}{{3x + 1}}} \right):\dfrac{{6{x^2} + 10x}}{{1 - 6x + 9{x^2}}}\\
 = \dfrac{{3x.\left( {3x + 1} \right) + 2x.\left( {1 - 3x} \right)}}{{\left( {1 - 3x} \right)\left( {3x + 1} \right)}}:\dfrac{{2x\left( {3x + 5} \right)}}{{{{\left( {3x - 1} \right)}^2}}}\\
 = \dfrac{{9{x^2} + 3x + 2x - 6{x^2}}}{{\left( {1 - 3x} \right)\left( {3x + 1} \right)}}.\dfrac{{{{\left( {3x - 1} \right)}^2}}}{{2x.\left( {3x + 5} \right)}}\\
 = \dfrac{{3{x^2} + 5x}}{{3x + 1}}.\dfrac{{1 - 3x}}{{2x.\left( {3x + 5} \right)}}\\
 = \dfrac{{x\left( {3x + 5} \right)}}{{3x + 1}}.\dfrac{{1 - 3x}}{{2x.\left( {3x + 5} \right)}}\\
 = \dfrac{{1 - 3x}}{{2.\left( {3x + 1} \right)}}
\end{array}\)