Đáp án:

$\begin{array}{l}
B1)\\
a)\dfrac{3}{4}:\left( {\dfrac{2}{3} - \dfrac{5}{9}} \right) + \dfrac{9}{4}\\
 = \dfrac{3}{4}:\left( {\dfrac{1}{9}} \right) + \dfrac{9}{4}\\
 = \dfrac{{27}}{4} + \dfrac{9}{4}\\
 = \dfrac{{36}}{4}\\
 = 9\\
b)\dfrac{{45}}{9} - {\left( {\dfrac{1}{2} + {{\left( {\dfrac{1}{3} + {{\left( {\dfrac{1}{4}} \right)}^{ - 1}}} \right)}^{ - 1}}} \right)^{ - 1}}\\
 = \dfrac{{45}}{9} - {\left( {\dfrac{1}{2} + {{\left( {\dfrac{1}{3} + 4} \right)}^{ - 1}}} \right)^{ - 1}}\\
 = \dfrac{{45}}{9} - {\left( {\dfrac{1}{2} + {{\left( {\dfrac{{13}}{3}} \right)}^{ - 1}}} \right)^{ - 1}}\\
 = \dfrac{{45}}{9} - {\left( {\dfrac{1}{2} + \dfrac{3}{{13}}} \right)^{ - 1}}\\
 = \dfrac{{45}}{9} - {\left( {\dfrac{{13 + 3.2}}{{26}}} \right)^{ - 1}}\\
 = \dfrac{{45}}{9} - {\left( {\dfrac{{19}}{{26}}} \right)^{ - 1}}\\
 = \dfrac{{45}}{9} - \dfrac{{26}}{{19}}\\
 = \dfrac{{45.19 - 9.26}}{{9.19}}\\
 = \dfrac{{621}}{{171}}\\
c)\dfrac{{{{5.4}^{15}}{{.9}^9} - {{4.3}^{20}}{{.8}^9}}}{{{{5.2}^{10}}{{.6}^{19}} - {{7.2}^{29}}{{.27}^6}}}\\
 = \dfrac{{{{5.2}^{30}}{{.3}^{18}} - {{2.3}^{20}}{{.2}^{27}}}}{{{{5.2}^{10}}{{.2}^{19}}{{.3}^{19}} - {{7.2}^{29}}{{.3}^{18}}}}\\
 = \dfrac{{{2^{28}}{{.3}^{18}}.\left( {{{5.2}^2} - {3^2}} \right)}}{{{2^{29}}{{.3}^{18}}.\left( {5.3 - 7} \right)}}\\
 = \dfrac{{20 - 9}}{{2.\left( {15 - 7} \right)}}\\
 = \dfrac{{11}}{{16}}\\
B2)\\
a)2\left( {x - 1} \right) - 3\left( {2x + 2} \right) - 4\left( {2x + 3} \right) = 16\\
 \Leftrightarrow 2x - 2 - 6x - 6 - 8x - 12 = 16\\
 \Leftrightarrow  - 12x = 36\\
 \Leftrightarrow x =  - 3\\
Vậy\,x =  - 3\\
b)3\dfrac{1}{2}:\left| {2x - 1} \right| = \dfrac{{21}}{{22}}\\
 \Leftrightarrow \dfrac{7}{2}:\left| {2x - 1} \right| = \dfrac{{21}}{{22}}\\
 \Leftrightarrow \left| {2x - 1} \right| = \dfrac{7}{2}:\dfrac{{21}}{{22}}\\
 \Leftrightarrow \left| {2x - 1} \right| = \dfrac{7}{2}.\dfrac{{22}}{{21}} = \dfrac{{11}}{7}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = \dfrac{{11}}{7} \Leftrightarrow x = \dfrac{9}{7}\\
2x - 1 =  - \dfrac{{11}}{7} \Leftrightarrow x = \dfrac{{ - 2}}{7}
\end{array} \right.\\
Vậy\,x = \dfrac{9}{7};x = \dfrac{{ - 2}}{7}\\
c)Do:x + z = 2y\\
 \Leftrightarrow x + z - 2y = 0\\
 \Leftrightarrow 2x + 2z - 4y = 0\\
\dfrac{{2x - y}}{5} = \dfrac{{3y - 2z}}{{15}}\\
 = \dfrac{{2x - y - \left( {3y - 2z} \right)}}{{5 - 15}}\\
 = \dfrac{{2x + 2z - 4y}}{{ - 10}}\\
 = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x - y = 0\\
3y - 2z = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{y}{2}\\
z = \dfrac{3}{2}y
\end{array} \right.\\
Thay\,x + z - 2y = 0\\
 \Leftrightarrow \dfrac{y}{2} + \dfrac{3}{2}y - 2y = 0\\
 \Leftrightarrow y = 0\\
 \Leftrightarrow x = y = z = 0\\
Vậy\,x = y = z = 0
\end{array}$