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2 tháng 1 2018
a) \(\dfrac{1}{7}.\dfrac{-3}{8}+\dfrac{1}{7}.\dfrac{-13}{8}\)
\(=\dfrac{1}{7}.\left(\dfrac{-3}{8}+\dfrac{-13}{8}\right)\)
\(=\dfrac{1}{7}.\left(-2\right)\)
\(=\dfrac{-2}{7}\)
b) \(\dfrac{3}{5}.\dfrac{13}{46}-\dfrac{1}{10}.\dfrac{16}{23}\)
\(=\dfrac{39}{230}\) - \(\dfrac{8}{115}\)
\(=\dfrac{1}{10}\)
KT
2
KT
2
NX
24 tháng 8 2023
\(\dfrac{3}{7}.\dfrac{9}{36}-\dfrac{1}{14}.\dfrac{1}{13}\)
\(=\dfrac{3.9}{7.36}-\dfrac{1.1}{14.13}\)
\(=\dfrac{3}{28}-\dfrac{1}{182}=\dfrac{37}{364}\)
28 tháng 6 2016
a) \(\frac{3}{5}\cdot\frac{13}{46}-\frac{1}{10}\cdot\frac{16}{23}=\frac{39-16}{10\cdot23}=\frac{1}{10}\)
b) \(\frac{3}{7}\cdot\frac{9}{26}-\frac{1}{14}\cdot\frac{1}{13}=\frac{27-1}{14\cdot13}=\frac{2\cdot13}{2\cdot7\cdot13}=\frac{1}{7}\)
HB
2
31 tháng 10 2021
a, \(\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)=\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}.\left(-\dfrac{2}{3}\right)==\dfrac{3}{5}\left(-\dfrac{8}{3}-\dfrac{2}{3}\right)=\dfrac{3}{5}.\left(-\dfrac{10}{3}\right)=-2\)
b, \(-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)-\left(-\dfrac{21}{15}\right)=-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)+\dfrac{7}{5}=\dfrac{10}{7}+\dfrac{7}{5}=\dfrac{50+49}{35}=\dfrac{99}{35}\)
31 tháng 10 2021
a: \(=\dfrac{3}{5}\cdot\left(-\dfrac{8}{3}+\dfrac{-2}{3}\right)=\dfrac{3}{5}\cdot\dfrac{-10}{3}=-2\)
c: \(=\left(0.125\right)^{650}\cdot8^{102}\)
\(=\left(0.125\cdot8\right)^{102}\cdot\left(0.125\right)^{548}\)
\(=\dfrac{1}{8^{548}}\)
8 tháng 10 2023
2:
a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)
1:
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 10 2023
Bài 1:
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)
\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)
\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)
\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)
\(A=\dfrac{-16-1}{4}\)
\(A=-\dfrac{17}{4}\)
Bài 2:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)
\(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}\)
\(=\frac{27}{182}-\frac{1}{182}\)
\(=\frac{1}{7}\)
\(\frac{3}{5}.\frac{13}{46}-\frac{1}{10}.\frac{16}{23}\)
\(=\frac{6}{10}.\frac{13}{46}-\frac{1}{10}.\frac{16}{23}\)
\(=\frac{1}{10}.\frac{78}{46}-\frac{1}{10}.\frac{16}{32}\)
\(=\frac{1}{10}\left(\frac{78}{46}-\frac{32}{46}\right)\)
\(=\frac{1}{10}.1=\frac{1}{10}\)