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11 tháng 6 2019
\(a,\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+\frac{17}{-35}\)
=\(-\frac{5}{13}+\left(\frac{3}{17}+\frac{14}{17}\right)+\left(\frac{-18}{35}+\frac{-17}{35}\right)\)
= \(-\frac{5}{13}+1+\left(-1\right)\)
=\(-\frac{5}{13}\)
\(b,\frac{-3}{8}.\frac{1}{6}+\frac{3}{-8}.\frac{5}{6}+\frac{-10}{6}\)
=\(\frac{-3}{8}.\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{-10}{6}\)
=\(\frac{-3}{8}.1+\frac{-10}{6}\)
=\(-\frac{49}{24}\)
\(c,\frac{-4}{11}.\frac{5}{15}.\frac{11}{-4}\)
=\(\left(\frac{-4}{11}.\frac{11}{-4}\right).\frac{1}{3}\)
=\(1.\frac{1}{3}=\frac{1}{3}\)
\(d,\frac{13}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}\)
=\(\frac{13}{8}+\frac{1}{2}+\frac{-1}{8}\)
=\(\left(\frac{13}{8}+\frac{-1}{8}\right)+\frac{1}{2}\)
=\(\frac{3}{2}+\frac{1}{2}=2\)
\(e,\frac{-1}{2^2}-\left(-2\right)^2-5\)
=\(\frac{-1}{4}-4-5\)
=\(-\frac{37}{4}\)
\(f,\frac{121}{3}-\frac{5}{7}:\left(24-\frac{23}{57}\right)\)
=\(\frac{121}{3}-\frac{5}{7}:\frac{1345}{57}\)
=\(\frac{121}{3}-\frac{57}{1883}\)
\(\approx40,4\)
PT
4
9 tháng 6 2018
\(\left(6-2\frac{4}{5}\right)\cdot3\frac{1}{8}+1\frac{3}{8}\div\frac{1}{4}\)
\(=\left(4+\frac{4}{5}\right)\cdot\frac{25}{8}+\frac{11}{8}\div\frac{1}{4}\)
\(=\frac{24}{5}\cdot\frac{25}{8}+\frac{44}{8}\)
\(=15+\frac{11}{2}\)
\(=\frac{41}{2}\)
9 tháng 6 2018
\((6-2\frac{4}{5})\cdot3\frac{1}{8}+1\frac{3}{8}:\frac{1}{4}\)
=\((4+\frac{4}{5})\cdot\frac{25}{8}+\frac{11}{8}:\frac{1}{4}\)
=\((\frac{24}{5}\cdot\frac{25}{8})+(\frac{11}{8}:\frac{1}{4})\)
=\(15+5,5\)
=\(20,5\)
18 tháng 2 2021
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PV
1
21 tháng 4 2022
a, \(\dfrac{-3}{5}+\dfrac{5}{7}\)
\(=\dfrac{4}{35}\)
b, \(-5+\dfrac{1}{6}\)
= \(\dfrac{-29}{6}\)
c, \(\dfrac{1}{8}-\dfrac{1}{2}\)
= \(\dfrac{-3}{8}\)
c, \(\dfrac{5}{12}-\dfrac{-7}{6}\)
= \(\dfrac{19}{12}\)
d, \(-2.\dfrac{3}{7}\)
= \(\dfrac{-6}{7}\)
\(\frac{1}{6}+\frac{8}{3}-\frac{2}{2}=\frac{1}{6}+\frac{16}{6}-1=\frac{11}{6}\)
\(\frac{1}{6}+\frac{8}{3}-\frac{1}{2}\)
= \(\frac{1}{6}+\frac{8}{3}+\frac{-1}{2}\)
\(=\frac{1}{6}+\frac{16}{6}+\frac{-3}{6}\)
\(=\frac{1+16+\left(-3\right)}{6}\)
\(=\frac{14}{6}=\frac{7}{3}\)