KC
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TD
2
28 tháng 3 2017
\(\dfrac{5}{7}x-x=1\dfrac{1}{7}\\ < =>\dfrac{5}{7}x-x=\dfrac{8}{7}\\ < =>\left(\dfrac{5}{7}-\dfrac{7}{7}\right)x=\dfrac{8}{7}\\ < =>-\dfrac{2}{7}x=\dfrac{8}{7}\\ =>x=\dfrac{\dfrac{8}{7}}{-\dfrac{2}{7}}=-4\)
2 tháng 5 2017
\(\dfrac{1}{7}.2\dfrac{1}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\dfrac{7}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\left(\dfrac{7}{3}-\dfrac{59}{6}\right)+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{1}{7}.\dfrac{-15}{2}+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{-15}{14}+\dfrac{15}{14}\)
= 0
ND
1
2 tháng 5 2017
=\(\dfrac{6}{7}+\dfrac{5}{8}.\dfrac{1}{5}-\dfrac{3}{16}.\left(-4\right)\)
= \(\dfrac{6}{7}+\dfrac{5.1}{8.5}-\dfrac{3.\left(-4\right)}{16}\)= \(\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{-3}{4}\)= \(\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{-42}{56}\)
= \(\dfrac{48+7-\left(-42\right)}{56}\)=\(\dfrac{97}{56}\)
LK
25 tháng 4 2018
a,
= \(\dfrac{20}{100}.\dfrac{15}{36}-\)(\(\dfrac{2}{5}+\dfrac{2}{3}\)):\(\dfrac{6}{5}\)
= \(\dfrac{20.15}{100.36}\)- ( \(\dfrac{6}{15}+\dfrac{10}{15}\)): \(\dfrac{6}{5}\)
= \(\dfrac{1.5}{5.12}-\dfrac{16}{15}:\dfrac{6}{5}\)
= \(\dfrac{1.1}{1.12}-\dfrac{16.5}{10.6}\)
= \(\dfrac{1}{12}-\dfrac{16.1}{2.6}\)
= \(\dfrac{1}{12}\) - \(\dfrac{16}{12}\)
= \(\dfrac{1}{12}+\dfrac{\left(-16\right)}{12}\)
= \(\dfrac{15}{12}\) = \(\dfrac{5}{4}\)
b,
= \(\dfrac{6}{7}+\dfrac{5}{4}+\dfrac{\left(-4\right)}{7}+\dfrac{25}{4}\)
= (\(\dfrac{6}{7}+\dfrac{\left(-4\right)}{7}\)) + (\(\dfrac{5}{4}+\dfrac{25}{4}\))
= \(\dfrac{2}{7}+\dfrac{30}{4}\)
= \(\dfrac{2}{7}+\dfrac{15}{2}\)
= \(\dfrac{4}{14}+\dfrac{105}{14}\)
= \(\dfrac{109}{14}\)
9 tháng 3 2017
a) \(\dfrac{5}{3}+\dfrac{3}{-4}+\dfrac{7}{6}\) \(\left(MC:12\right)\)
\(=\dfrac{20}{12}+\dfrac{-9}{12}+\dfrac{14}{12}\)
\(=\dfrac{20+\left(-9\right)+14}{12}\)
\(=\dfrac{25}{12}\)
b) \(\dfrac{-1}{5}+\dfrac{5}{3}+\dfrac{-3}{2}\) \(\left(MC:30\right)\)
\(=\dfrac{-6}{30}+\dfrac{50}{30}+\dfrac{-45}{30}\)
\(=\dfrac{\left(-6\right)+50+\left(-45\right)}{30}\)
\(=\dfrac{-1}{30}\)
c) \(\dfrac{2}{7}+\dfrac{-7}{5}+\dfrac{-2}{35}\) \(\left(MC:35\right)\)
\(=\dfrac{10}{35}+\dfrac{-49}{35}+\dfrac{-2}{35}\)
\(=\dfrac{10+\left(-49\right)+\left(-2\right)}{35}\)
\(=\dfrac{-41}{35}\)
d) \(3+\dfrac{-7}{2}+\dfrac{-1}{5}\) \(\left(MC:10\right)\)
\(=\dfrac{30}{10}+\dfrac{-35}{10}+\dfrac{-2}{10}\)
\(=\dfrac{30+\left(-35\right)+\left(-2\right)}{10}\)
\(=\dfrac{-7}{10}\)
9 tháng 3 2017
a) \(\dfrac{5}{3}+\dfrac{3}{-4}+\dfrac{7}{6}\)
\(=\dfrac{5}{3}+\dfrac{-3}{4}+\dfrac{7}{6}\)
\(=\) \(\dfrac{20}{12}+\dfrac{-9}{12}+\dfrac{14}{12}\)
\(=\dfrac{11}{12}+\dfrac{14}{12}\)
\(=\dfrac{25}{12}\)
b) \(\dfrac{-1}{5}+\dfrac{5}{3}+\dfrac{-3}{2}\)
\(=\dfrac{-6}{30}+\dfrac{50}{30}+\dfrac{-45}{30}\)
\(=\dfrac{44}{30}+\dfrac{-45}{30}\)
\(=\dfrac{-1}{30}\)
c) \(\dfrac{2}{7}+\dfrac{-7}{5}+\dfrac{-2}{35}\)
\(=\dfrac{10}{35}+\dfrac{-49}{35}+\dfrac{-2}{35}\)
\(=\dfrac{-39}{35}+\dfrac{-2}{35}\)
\(=\dfrac{-41}{35}\)
d) \(3+\dfrac{-7}{2}+\dfrac{-1}{5}\)
\(=\dfrac{3}{1}+\dfrac{-7}{2}+\dfrac{-1}{5}\)
\(=\dfrac{30}{10}+\dfrac{-35}{10}+\dfrac{-2}{10}\)
\(=\dfrac{-5}{10}+\dfrac{-2}{10}\)
\(=\dfrac{-7}{10}\)
TT
18 tháng 4 2018
\(\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
\(=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8})}{5\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}\)
\(=1\)
NA
27 tháng 7 2017
a)\(\dfrac{-1}{4}\cdot13\dfrac{9}{11}-0,25\cdot6\dfrac{2}{11}\)
\(=\dfrac{-1}{4}\cdot\dfrac{152}{11}-\dfrac{1}{4}\cdot\dfrac{68}{11}\)
=\(\dfrac{1}{4}\cdot\left(\dfrac{-152}{11}-\dfrac{68}{11}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{-220}{11}=-5\)
24 tháng 5 2022
a: \(=\dfrac{-3}{7}+\dfrac{-9}{35}-\dfrac{2}{5}\)
\(=\dfrac{-15-9-14}{35}=\dfrac{-38}{35}\)
b: \(=\left(\dfrac{15}{24}-\dfrac{7}{12}\right)\cdot\dfrac{-12}{7}\)
\(=\dfrac{15-14}{24}\cdot\dfrac{-12}{7}=\dfrac{1}{24}\cdot\dfrac{-12}{7}=\dfrac{-1}{14}\)
c: \(=\dfrac{7}{5}\cdot\dfrac{15}{19}\cdot\dfrac{-8}{15}+\dfrac{7}{15}\)
\(=\dfrac{-56}{95}+\dfrac{7}{15}\)
\(=\dfrac{-7}{57}\)
1/3 . 6/(-7) = -6/21 = -2/7
\(\dfrac{1}{3}\). \(\dfrac{6}{-7}\) = \(\dfrac{ }{7}\)
\(\dfrac{2}{-7}\) = \(\dfrac{ }{7}\)
\(◻\) = \(\dfrac{2}{-7}\) x 7
\(◻\) = \(-2\)