Đáp án:
$\begin{array}{l}
1)\left( {\dfrac{3}{{25}} - 1,12} \right):\dfrac{3}{7} - \left[ {\left( {3\dfrac{1}{2} - 3\dfrac{2}{3}} \right):\dfrac{1}{{14}}} \right]\\
= \left( {\dfrac{3}{{25}} - \dfrac{{28}}{{25}}} \right).\dfrac{7}{3} - \left[ {\left( {\dfrac{7}{2} - \dfrac{{11}}{3}} \right).14} \right]\\
= \dfrac{{ - 25}}{{25}}.\dfrac{7}{3} - \dfrac{{ - 1}}{6}.14\\
= \dfrac{{ - 7}}{3} + \dfrac{7}{3}\\
= 0\\
2)\left( {0,125} \right).\left( { - 3,7} \right).{\left( { - 2} \right)^3}\\
= \dfrac{1}{8}.\dfrac{{ - 37}}{{10}}.\left( { - 8} \right)\\
= 3,7\\
3)\sqrt {36} .\sqrt {\dfrac{{25}}{{16}}} + \dfrac{1}{4}\\
= 6.\dfrac{5}{4} + \dfrac{1}{4}\\
= \dfrac{{30}}{4} + \dfrac{1}{4}\\
= \dfrac{{31}}{4}\\
4)\sqrt {\dfrac{4}{{81}}} :\sqrt {\dfrac{{25}}{{81}}} - 1\dfrac{2}{5}\\
= \dfrac{2}{9}:\dfrac{5}{9} - \dfrac{7}{5}\\
= \dfrac{2}{5} - \dfrac{7}{5}\\
= - 1\\
5)0,1.\sqrt {225} .\sqrt {\dfrac{1}{4}} \\
= \dfrac{1}{{10}}.15.\dfrac{1}{2}\\
= \dfrac{3}{4}
\end{array}$
Câu 6 giống câu 1
