Đáp án:

$\begin{array}{l}
B2)\\
\dfrac{{5 - x}}{4} = \dfrac{{ - 9}}{{x - 5}} = \dfrac{{ - 15}}{{2y}}\\
 \Leftrightarrow \left( {5 - x} \right).\left( {x - 5} \right) = 4.\left( { - 9} \right)\\
 \Leftrightarrow {\left( {x - 5} \right)^2} = 36\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 6 \Leftrightarrow x = 11\left( {tm} \right)\\
x - 5 =  - 6 \Leftrightarrow x =  - 1\left( {tm} \right)
\end{array} \right.\\
 + Khi:x = 11\\
 \Leftrightarrow \dfrac{{5 - 11}}{4} = \dfrac{{ - 15}}{{2y}}\\
 \Leftrightarrow y = \dfrac{{4.\left( { - 15} \right)}}{{2.\left( { - 6} \right)}} = 5\left( {tm} \right)\\
 + Khi:x =  - 1\\
 \Leftrightarrow \dfrac{{5 - \left( { - 1} \right)}}{4} = \dfrac{{ - 15}}{{2y}}\\
 \Leftrightarrow y =  - 5\left( {tm} \right)\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {11;5} \right);\left( { - 1; - 5} \right)} \right\}\\
16)Do:a + c = 2b + 1\\
 \Leftrightarrow a - 2b + c = 1\\
P = \dfrac{3}{{16}}a - \dfrac{3}{8}b + \dfrac{3}{{16}}c\\
 = \dfrac{3}{{16}}.\left( {a - 2b + c} \right)\\
 = \dfrac{3}{{16}}.1 = \dfrac{3}{{16}}\\
B2)\\
a)\dfrac{2}{3}\left( {x - 1} \right) + 0,5\left( {2 - x} \right) =  - 3\\
 \Leftrightarrow \dfrac{2}{3}x - \dfrac{2}{3} + 1 - \dfrac{1}{2}x =  - 3\\
 \Leftrightarrow \dfrac{2}{3}x - \dfrac{1}{2}x =  - 3 - 1 + \dfrac{2}{3}\\
 \Leftrightarrow \dfrac{1}{6}x = \dfrac{{ - 10}}{3}\\
 \Leftrightarrow x =  - 20\\
Vậy\,x =  - 20\\
b)3\left( {x - \dfrac{1}{2}} \right) - 2\left( {x + 1} \right) = 125\% x - 1\\
 \Leftrightarrow 3x - \dfrac{3}{2} - 2x - 2 = \dfrac{5}{4}x - 1\\
 \Leftrightarrow \dfrac{5}{4}x - x =  - \dfrac{3}{2} - 2 + 1\\
 \Leftrightarrow \dfrac{1}{4}x =  - \dfrac{5}{2}\\
 \Leftrightarrow x =  - 10\\
Vậy\,x =  - 1\\
c)\dfrac{1}{2}\left( {2x + 3} \right) - \dfrac{1}{3}\left( {5 - 3x} \right) = 1\dfrac{1}{2} - x + 7\\
 \Leftrightarrow x + \dfrac{3}{2} - \dfrac{5}{3} + x = \dfrac{3}{2} - x + 7\\
 \Leftrightarrow 3x = 7 + \dfrac{5}{3}\\
 \Leftrightarrow x = \dfrac{{26}}{3}\\
Vậy\,x = \dfrac{{26}}{3}\\
g)9\left( {x + 1} \right)\left( { - 1 - x} \right) =  - 4\\
 \Leftrightarrow  - {\left( {x + 1} \right)^2} = \dfrac{{ - 4}}{9}\\
 \Leftrightarrow {\left( {x + 1} \right)^2} = \dfrac{4}{9}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 1 = \dfrac{2}{3} \Leftrightarrow x =  - \dfrac{1}{3}\\
x + 1 =  - \dfrac{2}{3} \Leftrightarrow x = \dfrac{{ - 5}}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 1}}{3};x = \dfrac{{ - 5}}{3}\\
h){\left( { - 3x - \dfrac{3}{4}} \right)^3} = \dfrac{{ - 8}}{{27}}\\
 \Leftrightarrow  - 3x - \dfrac{3}{4} =  - \dfrac{2}{3}\\
 \Leftrightarrow 3x = \dfrac{{ - 3}}{4} + \dfrac{2}{3}\\
 \Leftrightarrow 3x = \dfrac{{ - 1}}{{12}}\\
 \Leftrightarrow x =  - \dfrac{1}{{36}}\\
Vậy\,x = \dfrac{{ - 1}}{{36}}\\
i)9{\left( {2 - \dfrac{1}{3}x} \right)^2} = 4\\
 \Leftrightarrow {\left( {2 - \dfrac{1}{3}x} \right)^3} = \dfrac{4}{9}\\
 \Leftrightarrow \left[ \begin{array}{l}
2 - \dfrac{1}{3}x = \dfrac{2}{3}\\
2 - \dfrac{1}{3}x = \dfrac{{ - 2}}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{3}x = \dfrac{4}{3} \Leftrightarrow x = 4\\
\dfrac{1}{3}x = 2 + \dfrac{2}{3} = \dfrac{7}{3} \Leftrightarrow x = 7
\end{array} \right.\\
Vậy\,x = 4;x = 7
\end{array}$