Đáp án: C-C-B
Giải thích các bước giải:
$\begin{array}{l}
4)\dfrac{{{{\left[ {{{\left( 4 \right)}^{ - 2}}:{{\left( {\dfrac{{ - 1}}{3}} \right)}^2}} \right]}^{\dfrac{1}{2}}}}}{{{{\left( { - \dfrac{1}{6}} \right)}^2}}} = \dfrac{{{{\left( {\dfrac{1}{{16}}:\dfrac{1}{9}} \right)}^{\dfrac{1}{2}}}}}{{\dfrac{1}{{36}}}}\\
= 36.\sqrt {\left( {\dfrac{1}{{16}}.9} \right)} \\
= 36.\sqrt {\dfrac{9}{{16}}} \\
= 36.\dfrac{3}{4}\\
= 27\\
\Leftrightarrow C\\
5)\\
\left( {\dfrac{5}{2} - \dfrac{1}{{1 - \dfrac{1}{2}}}} \right):\left( {\dfrac{1}{2} - \dfrac{{\dfrac{1}{2}}}{{1 + \dfrac{1}{2}}}} \right)\\
= \left( {\dfrac{5}{2} - \dfrac{1}{{\dfrac{1}{2}}}} \right):\left( {\dfrac{1}{2} - \dfrac{{\dfrac{1}{2}}}{{\dfrac{3}{2}}}} \right)\\
= \left( {\dfrac{5}{2} - 2} \right):\left( {\dfrac{1}{2} - \dfrac{1}{3}} \right)\\
= \dfrac{1}{2}:\dfrac{1}{6}\\
= \dfrac{1}{2}.6\\
= 3\\
\Leftrightarrow C\\
6)\dfrac{{2x - 3}}{5} < \dfrac{{x - 2}}{7} + \dfrac{1}{3}\\
\Leftrightarrow \dfrac{{21\left( {2x - 3} \right)}}{{105}} < \dfrac{{15.\left( {x - 2} \right) + 35}}{{105}}\\
\Leftrightarrow 42x - 63 < 15x - 30 + 35\\
\Leftrightarrow 27x < 35 - 30 + 63\\
\Leftrightarrow 27x < 68\\
\Leftrightarrow x < \dfrac{{68}}{{27}}\\
\Leftrightarrow B
\end{array}$
