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\(\frac{5}{7}\times\frac{1}{3}-\frac{5}{7}\times\frac{1}{4}-\frac{5}{7}\times\frac{1}{2}\)
\(=\frac{5}{7}\times\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{5}{7}\times\left(\frac{4}{12}-\frac{3}{12}-\frac{6}{12}\right)\)
\(=\frac{5}{7}\times\left(\frac{4-3-6}{12}\right)\)
\(=\frac{5}{7}\times\frac{-5}{12}\)
\(=\frac{5\times\left(-5\right)}{7\times12}\)
\(=\frac{-25}{84}\)
\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{2}\)
= \(\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right).1\)
\(=\frac{5}{7}.\frac{-5}{12}\)
\(=-\frac{25}{84}\)
\(a,\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)
\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}=\frac{-7}{25}-\frac{18}{25}=-1\)
\(b,\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.0=0\)
c)\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}=\frac{27}{5}.\frac{30}{7}+\frac{40}{7}.\frac{27}{5}=\frac{27}{5}.\left(\frac{30}{7}+\frac{40}{7}\right)\)
\(=\frac{27}{5}.10=27.2=54\)
\(d,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2=\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}-\frac{1}{4}\)
\(=\left(\frac{3}{4}-\frac{1}{4}\right)-\frac{3}{2}+\frac{6}{5}=\frac{1}{2}-\frac{3}{2}+\frac{6}{5}=-1+\frac{6}{5}=\frac{-5}{5}+\frac{6}{5}=\frac{1}{5}\)
\(25\%-2\frac{1}{3}+0,5.\frac{12}{5}\)
\(=\frac{1}{4}-\frac{7}{3}+\frac{1}{2}.\frac{12}{5}\)
\(=\frac{1}{4}-\frac{7}{3}+\frac{6}{5}\)
\(=\frac{1}{4}+\frac{-7}{3}+\frac{6}{5}\)
\(=\frac{15}{60}+\frac{-140}{60}+\frac{72}{60}\)
\(=\frac{-53}{60}\)
a) \(\frac{2}{3}-\frac{5}{7}\cdot\frac{14}{25}\)
\(=\frac{2}{3}-\frac{2}{5}=\frac{4}{15}\)
b) \(\frac{-2}{5}\cdot\frac{5}{8}+\frac{5}{8}\cdot\frac{3}{5}\)
\(=\frac{5}{8}\left(-\frac{2}{5}+\frac{3}{5}\right)\)
\(=\frac{5}{8}\cdot\frac{1}{5}=\frac{1}{8}\)
c) \(25\%-1\frac{1}{2}+0,5\cdot\frac{12}{5}\)
\(=0,25-1,5+0,5\cdot2,4\)
\(=0,25-1,5+1,2=-0,05\)
a) \(\frac{2}{3}-\frac{5}{7}\cdot\frac{14}{25}=\frac{2}{3}-\frac{2}{5}=\frac{10-6}{15}=\frac{4}{15}\)
b) \(-\frac{2}{5}\cdot\frac{5}{8}+\frac{5}{8}\cdot\frac{3}{5}=\frac{5}{8}\cdot\left(-\frac{2}{5}+\frac{3}{5}\right)=\frac{5}{8}\cdot\frac{1}{5}=\frac{1}{8}\)
c) \(25\%-1\frac{1}{2}+0,5\cdot\frac{12}{5}=\frac{1}{4}-\frac{3}{2}+\frac{1}{2}\cdot\frac{12}{5}=\frac{1}{4}-\frac{3}{2}+\frac{6}{5}=\frac{5-30+24}{20}=-\frac{1}{20}\)
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Bài 1:
\(g.\frac{5}{11}+\frac{6}{11}=\frac{5+6}{11}=\frac{11}{11}=1\)\(\)
\(e.\frac{-17}{25}.\frac{20}{33}+\frac{-17}{25}.\frac{13}{33}+\frac{-3}{25}=\frac{-17}{25}.\left(\frac{20}{33}+\frac{13}{33}\right)+\frac{-3}{25}\)
\(=\frac{-17}{25}.1+\frac{-3}{25}=\frac{-17}{25}+\frac{-3}{25}=\frac{-17-3}{25}=\frac{-20}{25}=\frac{-4}{5}\)
\(d.\frac{5}{7}.\frac{19}{23}+\frac{5}{7}.\frac{5}{23}-\frac{5}{7}.\frac{1}{23}=\frac{5}{7}\left(\frac{19}{23}+\frac{5}{23}-\frac{1}{23}\right)\)
\(=\frac{5}{7}\left(\frac{19+5-1}{23}\right)=\frac{5}{7}.1=\frac{5}{7}\)
\(c.\left(-11\right).\frac{9}{22}=\frac{\left(-11\right).9}{22}=\frac{-99}{22}=\frac{-9}{2}\)
\(b.\frac{5}{6}-\frac{1}{8}=\frac{5.4}{6.4}-\frac{1.3}{8.3}=\frac{20}{24}-\frac{3}{24}=\frac{17}{24}\)
\(a.\frac{2}{3}+\frac{1}{5}-\frac{1}{6}=\frac{2.10}{3.10}+\frac{1.6}{5.6}-\frac{1.5}{6.5}=\frac{20}{30}+\frac{6}{30}-\frac{5}{30}\)
\(=\frac{20+6-5}{30}=\frac{21}{30}=\frac{7}{10}\)
Bài 2:
\(a.\frac{3}{4}+x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{3}{4}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{9}{12}\)
\(\Leftrightarrow x=\frac{2}{12}=\frac{1}{6}\)
\(b.x-\frac{11}{12}=0,5\)
\(\Leftrightarrow x=\frac{1}{2}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{6}{12}+\frac{11}{12}\)
\(\Leftrightarrow x=\frac{17}{12}\)
a) \(x+\frac{5}{12}=-1\frac{2}{7}\)
\(\Leftrightarrow x+\frac{5}{12}=\frac{-9}{7}\)
\(\Leftrightarrow x=\frac{-143}{84}\)
Vậy ...
b) \(4\frac{1}{2}x:\frac{5}{12}=0,5\)
\(\Leftrightarrow\frac{9}{2}x=\frac{5}{24}\)
\(\Leftrightarrow x=\frac{5}{108}\)
vậy...
c) \(7,5.1\frac{3}{4}x=6\frac{2}{5}\)
\(\Leftrightarrow\frac{105}{8}x=\frac{32}{5}\)
\(\Leftrightarrow x=\frac{256}{525}\)
Vậy ...
\(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
\(=\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
\(=-1+1-\frac{1}{2}=0-\frac{1}{2}\)
\(=\frac{-1}{2}\)
\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5=2,5\)
\(\frac{13}{25}+\frac{4}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
= \(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
= \(-1+1-\frac{1}{2}=-\frac{1}{2}\)
\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
=\(\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
= \(1+1+0,5=2,5\)
25% - 1 1/2 + 0,5 . 12/5
= 1/4 - 3/2 + 1/2 . 12 /5
= 1/4 - 3/2 + 6/5
= -5/4 + 6/5
= -1/20
25% - 1 1/2 + 0.5 x 12/5
= 0.25 - 1.5 + 0.5 x 2.4
= 0.25 - 1.5 + 1.2
= 0.25 - 2.7
= -2.45