Đáp án:
$\begin{array}{l}
119)\\
C = \dfrac{{12 - \sqrt {15.135} + {{\left( {\sqrt {31} } \right)}^2}}}{{\sqrt {\dfrac{{80}}{{45}}} - \dfrac{{10}}{{{{\left( {\sqrt 3 } \right)}^2}}}}}\\
= \dfrac{{12 - \sqrt {15.15.9} + 31}}{{\sqrt {\dfrac{{16}}{9}} - \dfrac{{10}}{3}}}\\
= \dfrac{{12 - 15.3 + 31}}{{\dfrac{4}{3} - \dfrac{{10}}{3}}}\\
= \dfrac{{ - 2}}{{\dfrac{{ - 6}}{3}}} = \dfrac{{ - 2}}{{ - 2}} = 1\\
120)\\
x = \dfrac{{ - 25}}{{12}};y = - 3\\
\Rightarrow D = 5.\left( {\dfrac{{ - 25}}{{12}}} \right) - 7.\sqrt {\dfrac{{ - 25}}{{12}}.\left( { - 3} \right)} + 2.\left( { - 3} \right)\\
= \dfrac{{ - 125}}{{12}} - 7.\sqrt {\dfrac{{25}}{4}} - 6\\
= \dfrac{{ - 125}}{{12}} - 6 - 7.\dfrac{5}{2}\\
= \dfrac{{ - 125 - 6.12 - 35.6}}{{12}}\\
= \dfrac{{407}}{{12}}\\
122)\\
x - \dfrac{7}{{10}}:{\left( {0,4} \right)^2} = {\left( {\dfrac{3}{2}} \right)^2} + \dfrac{5}{6}.\left( { - 7,95} \right)\\
\Rightarrow x - \dfrac{7}{{10}}:\dfrac{4}{{25}} = \dfrac{9}{4} + \dfrac{5}{6}.\dfrac{{ - 159}}{{20}}\\
\Rightarrow x - \dfrac{{35}}{8} = \dfrac{9}{4} - \dfrac{{53}}{8}\\
\Rightarrow x = \dfrac{9}{4} - \dfrac{{53}}{8} + \dfrac{{35}}{8}\\
\Rightarrow x = \dfrac{9}{4} - \dfrac{9}{4}\\
\Rightarrow x = 0\\
Vậy\,x = 0
\end{array}$
