\(\frac{-8}{15}+\frac{5}{18}-\frac{17}{12}\)
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HQ
Hà Quang Minh
Giáo viên
2 tháng 11 2023
a) $\frac{1}{6}:\frac{3}{7} = \frac{1}{6} \times \frac{7}{3} = \frac{7}{{18}}$
b) $\frac{5}{{12}}:\frac{1}{4} = \frac{5}{{12}} \times \frac{4}{1} = \frac{{5 \times 4}}{{12 \times 1}} = \frac{{5 \times 4}}{{4 \times 3 \times 1}} = \frac{5}{3}$
c) $\frac{4}{{15}}:\frac{8}{3} = \frac{4}{{15}} \times \frac{3}{8} = \frac{{4 \times 3}}{{15 \times 8}} = \frac{{4 \times 3}}{{5 \times 3 \times 4 \times 2}} = \frac{1}{{10}}$
d) $\frac{{18}}{5}:\frac{9}{{10}} = \frac{{18}}{5} \times \frac{{10}}{9} = \frac{{18 \times 10}}{{5 \times 9}} = \frac{{9 \times 2 \times 5 \times 2}}{{5 \times 9}} = 4$
BB
20 tháng 1 2017
a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
TN
23 tháng 7 2016
\(E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)-\left(\frac{12}{17}-\frac{5}{18}+\frac{7}{5}\right)\)
\(=\frac{1}{13}+\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}-\frac{12}{17}+\frac{5}{18}-\frac{7}{5}\)
\(=\left(\frac{1}{13}-\frac{1}{13}\right)+\left(\frac{-5}{18}+\frac{5}{18}\right)+\left(\frac{12}{17}-\frac{12}{17}\right)-\frac{7}{5}\)
\(=0+0+0-\frac{7}{5}\)
\(=-\frac{7}{5}\)
23 tháng 7 2016
\(E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)-\left(\frac{12}{17}-\frac{5}{18}+\frac{7}{5}\right)\)
\(=\frac{1}{13}+\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}-\frac{-12}{17}+\frac{5}{18}-\frac{7}{5}\)
\(=\left(\frac{1}{13}-\frac{1}{13}\right)+\left(\frac{-5}{18}+\frac{5}{18}\right)+\left(\frac{12}{17}-\frac{-12}{17}\right)-\frac{7}{5}\)
\(=0+0+\frac{24}{17}-\frac{7}{5}\)
\(=\frac{24}{17}-\frac{7}{5}\)
\(=\frac{120}{85}-\frac{119}{85}=\frac{1}{85}\)
1 tháng 8 2017
\(A=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(-\frac{20}{31}-\frac{11}{31}\right)+-\frac{4}{9}\)
\(A=1+\left(-1\right)+\frac{-4}{9}\)
\(A=0+\frac{-4}{9}=\frac{-4}{9}\)
\(B=\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{-2}{3}\)
\(B=-1+1+\frac{-2}{3}\)
\(B=\frac{-2}{3}\)
\(\frac{-8}{15}+\frac{5}{18}-\frac{17}{12}=\frac{-96+50-255}{180}=\frac{-301}{180}\)
=\(\frac{-23}{90}-\frac{17}{12}\)
=\(-\frac{59}{180}\)