Đáp án:

$\begin{array}{l}
1)a)\dfrac{5}{6} - \dfrac{1}{2} + \dfrac{7}{{12}}\\
 = \dfrac{{5.2 - 6 + 7}}{{12}}\\
 = \dfrac{{11}}{{12}}\\
b)B = \left( {\dfrac{5}{9} - \dfrac{3}{7} + 3\dfrac{2}{3}} \right) + \left( {\dfrac{{17}}{7} - \dfrac{{14}}{4}} \right)\\
 = \dfrac{5}{9} - \dfrac{3}{7} + \dfrac{{11}}{3} + \dfrac{{17}}{7} - \dfrac{7}{2}\\
 = \dfrac{5}{9} + \dfrac{{33}}{9} + \dfrac{{17}}{7} - \dfrac{3}{7} - \dfrac{7}{2}\\
 = \dfrac{{38}}{9} + \dfrac{{14}}{7} - \dfrac{7}{2}\\
 = \dfrac{{38}}{9} + 2 - \dfrac{7}{2}\\
 = \dfrac{{38.2 + 2.9.2 - 7.9}}{{18}}\\
 = \dfrac{{76 + 36 - 63}}{{18}}\\
 = \dfrac{{48}}{{18}} = \dfrac{8}{3}\\
C = \dfrac{6}{{13}}.\dfrac{3}{{11}} + \dfrac{7}{{11}}.\dfrac{6}{{13}} - \dfrac{{21}}{{11}}:\dfrac{{13}}{6}\\
 = \dfrac{6}{{13}}.\dfrac{3}{{11}} + \dfrac{6}{{13}}.\dfrac{7}{{11}} - \dfrac{{21}}{{11}}.\dfrac{6}{{13}}\\
 = \dfrac{6}{{13}}.\left( {\dfrac{3}{{11}} + \dfrac{7}{{11}} - \dfrac{{21}}{{11}}} \right)\\
 = \dfrac{6}{{13}}.\dfrac{{ - 11}}{{11}}\\
 =  - \dfrac{6}{{13}}\\
2)a)\dfrac{{7 - x}}{{3 - 2x}} + \dfrac{1}{4} =  - \dfrac{5}{6}\\
 \Rightarrow \dfrac{{7 - x}}{{3 - 2x}} = \dfrac{{ - 5}}{6} - \dfrac{1}{4} = \dfrac{{ - 13}}{{12}}\\
 \Rightarrow  - 13.\left( {3 - 2x} \right) = 12.\left( {7 - x} \right)\\
 \Rightarrow  - 39 + 26x = 84 - 12x\\
 \Rightarrow 26x + 12x = 84 + 39\\
 \Rightarrow 38x = 123\\
 \Rightarrow x = \dfrac{{123}}{{38}}\\
\text{Vậy}\,x = \dfrac{{123}}{{38}}\\
b)\dfrac{{ - 10}}{9} + {\left( {\dfrac{{5x + 1}}{3}} \right)^2} = \dfrac{{14}}{5} - \dfrac{9}{5} = 1\\
 \Rightarrow {\left( {\dfrac{{5x + 1}}{3}} \right)^2} = \dfrac{{19}}{9}\\
 \Rightarrow \left[ \begin{array}{l}
\dfrac{{5x + 1}}{3} = \dfrac{{\sqrt {19} }}{3}\\
\dfrac{{5x + 1}}{3} = \dfrac{{ - \sqrt {19} }}{3}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt {19}  - 1}}{5}\\
x = \dfrac{{ - \sqrt {19}  - 1}}{5}
\end{array} \right.\\
c)3x =  - 5y = 2z\\
 \Rightarrow \dfrac{{3x}}{{30}} = \dfrac{{ - 5y}}{{30}} = \dfrac{{2z}}{{30}}\\
 \Rightarrow \dfrac{x}{{10}} = \dfrac{y}{{ - 6}} = \dfrac{z}{{15}}\\
 = \dfrac{{2x}}{{20}} = \dfrac{{3y}}{{ - 18}} = \dfrac{{2x + 3y}}{{20 - 18}} = \dfrac{{ - 1}}{2}\\
 \Rightarrow \left\{ \begin{array}{l}
x =  - \dfrac{1}{2}.10 =  - 5\\
y =  - \dfrac{1}{2}.\left( { - 6} \right) = 3\\
z =  - \dfrac{1}{2}.15 =  - \dfrac{{15}}{2}
\end{array} \right.\\
\text{Vậy}\,x =  - 5;y = 3;z =  - \dfrac{{15}}{2}
\end{array}$