a,\(\frac{5}{6}+\left\{\frac{-1}{2}\right\}+\frac{3}{4}\)
b,\(\left\{0,75-\frac{1}{3}\right\}:\frac{7}{15}\)
c, \(\frac{7}{12}-\frac{3}{4}.\frac{5}{6}\)
d,\(\left\{2\frac{1}{3}+1\frac{3}{4}\right\}.\frac{12}{13}\)
K
Khách
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2 tháng 7 2020
Bài 1:
a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=\frac{-2}{2}=-1\)
b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)
\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)
\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)
\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)
c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)
\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)
d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)
\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)
\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)
e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)
\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)
\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)
\(=\frac{11}{12}\)
f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)
\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)
\(=\frac{1}{5}\)
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)
\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)
\(a)\frac{5}{6}+\left(\frac{-1}{2}\right)+\frac{3}{4}\)
\(=\frac{1}{3}+\frac{3}{4}\)
\(=\frac{13}{12}\)
\(b)\left(0,75-\frac{1}{3}\right):\frac{7}{15}\)
\(=\left(\frac{3}{4}-\frac{1}{3}\right).\frac{15}{7}\)
\(=\frac{5}{12}.\frac{15}{7}\)
\(=\frac{25}{28}\)
\(c)\frac{7}{12}-\frac{3}{4}.\frac{5}{6}\)
\(=\frac{7}{12}-\frac{5}{8}\)
\(=\frac{-1}{24}\)
\(d)\left(2\frac{1}{3}+1\frac{3}{4}\right).\frac{12}{13}\)
\(=\left(\frac{7}{3}+\frac{7}{4}\right).\frac{12}{13}\)
\(=\frac{49}{12}.\frac{12}{13}\)
\(=\frac{49}{13}\)
a)\(\frac{5}{6}+\frac{-1}{2}+\frac{3}{4}=\frac{10}{12}-\frac{6}{12}+\frac{9}{12}=\frac{10-6+9}{12}=\frac{13}{12}\)
b)\(\left\{0,75-\frac{1}{3}\right\}:\frac{7}{15}=\left\{\frac{3}{4}-\frac{1}{3}\right\}.\frac{15}{7}=\left\{\frac{9}{12}-\frac{4}{12}\right\}.\frac{15}{7}=\frac{5}{12}.\frac{15}{7}=\frac{75}{84}\)
c)\(\frac{7}{12}-\frac{3}{4}.\frac{5}{6}=\frac{7}{12}-\frac{5}{8}=\frac{14}{24}-\frac{15}{24}=\frac{-1}{24}\)
d)\(\left\{2\frac{1}{3}+1\frac{3}{4}\right\}.\frac{12}{13}=\left\{\frac{7}{3}+\frac{7}{4}\right\}.\frac{12}{13}=\left\{\frac{28}{12}+\frac{21}{12}\right\}.\frac{12}{13}=\frac{39}{12}.\frac{12}{13}=3\)