Thực hiện phép tính:
a,\(\frac{15}{33}+\frac{7}{20}+\frac{18}{33}+\frac{13}{20}\)
b, \(2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{21}\right)\)
c, \(\left(\frac{-1}{2}\right)^3+\frac{1}{2}:5\)
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HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
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)
\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)
T
14 tháng 4 2019
\(A=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{3}{8}+\frac{-6}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{-3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\left(\frac{-36}{24}+\frac{56}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(A=\frac{5}{6}:\frac{5}{6}+\frac{1}{2}\)
\(A=\frac{5}{6}\times\frac{6}{5}+\frac{1}{2}\)
\(A=1+\frac{1}{2}\)
\(A=\frac{1}{1}+\frac{1}{2}=\frac{2}{2}+\frac{1}{2}\)
\(A=\frac{3}{2}\)
\(a.\)
\(\frac{15}{33}+\frac{7}{20}+\frac{18}{33}+\frac{13}{20}\)
\(=\left(\frac{15}{33}+\frac{18}{33}\right)+\left(\frac{13}{20}+\frac{7}{20}\right)\)
\(=\frac{33}{33}+\frac{20}{20}\)
\(=1+1=2\)
\(b.\)
\(2\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{21}\right)\)
\(=\frac{5}{2}+\frac{4}{7}:\left(-\frac{8}{21}\right)\)
\(=\frac{5}{2}+\frac{4}{7}.\left(-\frac{21}{8}\right)\)
\(=\frac{5}{2}+\frac{1}{1}.\left(-\frac{3}{2}\right)\)
\(=\frac{5}{2}-\frac{3}{2}\)
\(=1\)
\(c.\)
\(\left(-\frac{1}{2}\right)^3+\frac{1}{2}:5\)
\(=-\frac{1}{8}+\frac{1}{2}.\frac{1}{5}\)
\(=-\frac{1}{8}+\frac{1}{10}\)
\(=-\frac{1}{40}\)
a) \(\frac{15}{33}+\frac{7}{20}+\frac{18}{33}+\frac{13}{20}=\left(\frac{15}{33}+\frac{18}{33}\right)+\left(\frac{7}{20}+\frac{13}{20}\right)\) = 1 + 1 = 2
b) \(2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{21}\right)=\frac{5}{2}+\frac{4}{7}:\left(\frac{-8}{21}\right)=\frac{5}{2}+\frac{-3}{2}\) = 1
c) \(\left(\frac{-1}{2}\right)\)3 + \(\frac{1}{2}\) : 5 = \(\frac{-1}{8}+\frac{1}{10}\) = \(\frac{-1}{40}\)
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