Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\begin{array}{l}0,75 - \frac{5}{6} + 1\frac{1}{2} = \frac{3}{4} - \frac{5}{6} + \frac{3}{2}\\ = \frac{9}{{12}} - \frac{{10}}{{12}} + \frac{{18}}{{12}} = \frac{{17}}{{12}}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ - 8}}{{21}}} \right) + \left( { - 0,4} \right) = \frac{3}{7} + \frac{4}{{15}} - \frac{8}{{21}} - \frac{2}{5}\\ = \left( {\frac{3}{7} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{2}{5}} \right)\\ = \left( {\frac{9}{{21}} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{6}{{15}}} \right)\\ = \frac{1}{{21}} + \left( {\frac{{ - 2}}{{15}}} \right)\\ = \frac{5}{{105}} - \frac{{14}}{{105}}\\ = \frac{{ - 9}}{{105}} = \frac{{ - 3}}{{35}}\end{array}\)
c)
\(\begin{array}{l}0,625 + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ - 5}}{7}} \right) + 1\frac{2}{3}\\ = \frac{5}{8} + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} - \frac{5}{7} + \frac{5}{3}\\ = \left( {\frac{5}{8} + \frac{3}{8}} \right) + \left( {\frac{{ - 2}}{7} - \frac{5}{7}} \right) + \frac{5}{3}\\ = 1 - 1 + \frac{5}{3} = \frac{5}{3}\end{array}\)
d)
\(\begin{array}{l}\left( { - 3} \right).\left( {\frac{{ - 38}}{{21}}} \right).\left( {\frac{{ - 7}}{6}} \right).\left( { - \frac{3}{{19}}} \right)\\ = \frac{{ - 3.\left( { - 38} \right).\left( { - 7} \right).\left( { - 3} \right)}}{{21.6.19}}\\ = \frac{{3.38.7.3}}{{21.6.19}}\\ = \frac{{3.2.19.7.3}}{{3.7.3.2.19}}\\ = 1\end{array}\)
e)
\(\begin{array}{l}\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5} = \left( {\frac{{11}}{{18}}.\frac{9}{{22}}} \right).\frac{8}{5}\\ = \frac{{11.9.4.2}}{{9.2.2.11.5}} = \frac{2}{5}\end{array}\)
g)
\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{1}{8}.\frac{3}{5} = \frac{{1.3}}{{8.5}} = \frac{3}{{40}}\)
b)\(\frac{{ - 6}}{7}:\left( { - \frac{5}{3}} \right) = \frac{{ - 6}}{7}.\frac{{ - 3}}{5} = \frac{{18}}{{35}}\)
c)\(0,6.\left( { - 0,15} \right) = \frac{6}{{10}}.\frac{{ - 15}}{{100}} = \frac{{ - 90}}{{1000}} = \frac{{ - 9}}{{100}}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\begin{array}{l}\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right)\\ = \left( {\frac{{ - 3}}{7}} \right) + \frac{5}{6} - \frac{4}{7}\\ = \left[ {\left( {\frac{{ - 3}}{7}} \right) - \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\= - 1 + \frac{5}{6}\\ = \frac{{ - 1}}{6}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} - \frac{2}{3} - \frac{1}{5}\\ = (\frac{3}{5} - \frac{1}{5}) - \frac{2}{3}\\ = \frac{2}{5} - \frac{2}{3}\\ = \frac{6}{{15}} - \frac{{10}}{{15}}\\ = \frac{{ - 4}}{{15}}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{3}} \right) + 1} \right] - \left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{3}} \right) + 1 - \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ - 1}}{3} - \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= - 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)
d)
\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ = - \frac{3}{4}\end{array}\).
\(\begin{array}{l}a)\left[ {{{\left( {\dfrac{3}{7}} \right)}^4}.{{\left( {\dfrac{3}{7}} \right)}^5}} \right]:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{4 + 5}}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^9}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{9-7}}\\= {\left( {\dfrac{3}{7}} \right)^2}\\b)\left[ {{{\left( {\dfrac{7}{8}} \right)}^5}:{{\left( {\dfrac{7}{8}} \right)}^4}} \right].\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^{5 - 4}}.\left( {\dfrac{7}{8}} \right)\\ = \left( {\dfrac{7}{8}} \right).\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^2}\\c)\left[ {{{\left( {0,6} \right)}^3}.{{\left( {0,6} \right)}^8}} \right]:\left[ {{{\left( {0,6} \right)}^7}.{{\left( {0,6} \right)}^2}} \right]\\ = {\left( {0,6} \right)^{3 + 8}}:{\left( {0,6} \right)^{7 + 2}}\\ = {\left( {0,6} \right)^{11}}:{\left( {0,6} \right)^9}\\ = {\left( {0,6} \right)^{11-9}}\\={\left( {0,6} \right)^2}.\end{array}\)