Đáp án:

$-$

Giải thích các bước giải:

a) `12/25*23/7-12/25*12/7`

`=12/25*(23/7-12/7)`

`=12/25*11/7`

`=132/175`

b) `13 2/7:(-8)/9+2 5/7:(-8)/9`

`=93/7*(-9)/8+19/7*(-9)/8`

`=(-9)/8*(93/7+19/7)`

`=(-9)/8*16`

`=`$-18$

c) $\dfrac{-6}{11}\cdot\dfrac{7}{10}\cdot\dfrac{11}{-6}\cdot(-20)\\=\dfrac{-6}{11}:\dfrac{-6}{11}\cdot\dfrac{7}{10}\cdot(-20)\\=1\cdot\dfrac{7}{10}\cdot10.(-2)\\=7.(-2)\\=-14$

d) $-1\dfrac12:\dfrac34\cdot\left(-4\dfrac12\right)\\=\dfrac{-3}2:\dfrac34\cdot\dfrac{-9}{2}\\=-2\cdot\dfrac{-9}2\\=9$

e) $2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}}\\=2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{\dfrac52}}}\\=2+\dfrac{1}{2+\dfrac{1}{2+\dfrac25}}\\=2+\dfrac{1}{2+\dfrac{1}{\dfrac{12}{5}}}\\=2+\dfrac{1}{2+\dfrac{5}{12}}\\=2+\dfrac{1}{\dfrac{29}{12}}\\=2+\dfrac{12}{29}\\=\dfrac{70}{29}$

Đáp án:

$\begin{array}{l}
a)\dfrac{{12}}{{25}}.\dfrac{{23}}{7} - \dfrac{{12}}{{25}}.\dfrac{{12}}{7}\\
 = \dfrac{{12}}{{25}}.\left( {\dfrac{{23}}{7} - \dfrac{{12}}{7}} \right)\\
 = \dfrac{{12}}{{25}}.\dfrac{{11}}{7}\\
 = \dfrac{{132}}{{175}}\\
b)13\dfrac{2}{7}:\left( {\dfrac{{ - 8}}{9}} \right) + 2\dfrac{5}{7}:\left( {\dfrac{{ - 8}}{9}} \right)\\
 = \left( {13\dfrac{2}{7} + 2\dfrac{5}{7}} \right):\left( {\dfrac{{ - 8}}{9}} \right)\\
 = \left( {13 + \dfrac{2}{7} + 2 + \dfrac{5}{7}} \right).\dfrac{{ - 9}}{8}\\
 = \left( {15 + 1} \right).\dfrac{{ - 9}}{8}\\
 = 16.\dfrac{{ - 9}}{8}\\
 =  - 18\\
c)\left( { - \dfrac{6}{{11}}} \right).\dfrac{7}{{10}}.\left( {\dfrac{{11}}{{ - 6}}} \right).\left( { - 20} \right)\\
 =  - 14\\
d)\left( { - 1\dfrac{1}{2}} \right):\dfrac{3}{4}.\left( { - 4\dfrac{1}{2}} \right)\\
 = \dfrac{{ - 3}}{2}.\dfrac{4}{3}.\dfrac{{ - 9}}{2}\\
 = 9\\
e)2 + \dfrac{1}{{2 + \dfrac{1}{{2 + \dfrac{1}{{2 + \dfrac{1}{2}}}}}}}\\
 = 2 + \dfrac{1}{{2 + \dfrac{1}{{2 + \dfrac{1}{{\dfrac{5}{2}}}}}}}\\
 = 2 + \dfrac{1}{{2 + \dfrac{1}{{2 + \dfrac{2}{5}}}}}\\
 = 2 + \dfrac{1}{{2 + \dfrac{1}{{\dfrac{{12}}{5}}}}}\\
 = 2 + \dfrac{1}{{2 + \dfrac{5}{{12}}}}\\
 = 2 + \dfrac{1}{{\dfrac{{29}}{12}}}\\
 = 2 + \dfrac{12}{{29}}\\
 = \dfrac{{70}}{{29}}
\end{array}$