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\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)
\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)
\(=\frac{7}{12}+\frac{31}{12}\)
\(=\frac{38}{12}=\frac{19}{6}\)
\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)
\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\frac{2}{13}\)
\(=\frac{-10}{117}\)
\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)
\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)
\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)
\(=\frac{4}{5}\cdot\frac{13}{7}\)
\(=\frac{52}{35}\)
a)7/12.6/11+7/12.5/11-2.7/12
=7/12(6/11+5/11-2)
=7/12(1-2)
=7/12.(-1)
=-7/12
1) A = \(\frac{-15}{19}.\frac{23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\frac{-23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\left(\frac{-23}{37}+\frac{14}{37}\right)=\frac{15}{19}.\frac{-9}{37}=\frac{-135}{703}\)
a . 7/12 . 6/11 + 7/12 . 5/11 - 2 7/12
= 7/12 . ( 6/11 + 5/11 ) - 31/12
= 7/12 . 1 - 31/12
= 7/12 - 31/12
= -2
b . -5/9 . -6/13 + 5/-9 . -5/13 - 5/9
= -5/9 . ( -6/13 + -5/13 ) - 5/9
= -5/9 . ( -1 ) -5/9
= 5/9 - 5/9
= 0
a) \(\frac{2}{5}+\left(\frac{3}{11}+\frac{-2}{5}\right)\)
\(=\frac{2}{5}+\frac{3}{11}+\frac{-2}{5}\)
\(=\left(\frac{2}{5}+\frac{-2}{5}\right)+\frac{3}{11}\)
\(=0+\frac{3}{11}\)
\(=\frac{3}{11}\)
b) \(\frac{-5}{7}\cdot\frac{4}{13}+\frac{-5}{7}\cdot\frac{9}{13}+\frac{-2}{7}\)
\(=\frac{-5}{7}\left(\frac{4}{13}+\frac{9}{13}\right)+\frac{-2}{7}\)
\(=\frac{-5}{7}\cdot1+\frac{-2}{7}\)
\(=\frac{-5}{7}+\frac{-2}{7}\)
\(=-1\)
=))
a) 2/5 + (3/11 -2/5 ) = 2/5 - 2/5 + 3/11 = 3/11
b) -5/7*4/13+(-5/7 ) *9/13 + (-2/7) =-5/7( 4/13 + 9/13 ) -2/7 = -1
Bài làm
a) \(-\frac{3}{7}+\frac{3}{4}:\frac{3}{14}\)
= \(-\frac{3}{7}+\frac{3}{4}.\frac{14}{3}\)
= \(-\frac{3}{7}+\frac{7}{2}\)
\(=-\frac{7}{14}+\frac{49}{14}\)
\(=\frac{42}{14}=3\)
b) \(5-\frac{7}{39}:\frac{7}{13}+\frac{8}{9}:4\)
\(=5=\frac{7}{39}.\frac{13}{7}+\frac{8}{9}.\frac{1}{4}\)
\(=5-\frac{1}{3}+\frac{2}{9}\)
\(=\frac{45}{9}-\frac{3}{9}+\frac{2}{9}\)
\(=\frac{44}{9}\)
c) \(\left(\frac{5}{12}:\frac{11}{6}+\frac{5}{12}:\frac{11}{5}\right)-\frac{-7}{12}\)
\(=\left(\frac{5}{12}.\frac{6}{11}+\frac{5}{12}.\frac{5}{11}\right)+\frac{7}{12}\)
\(=\left[\frac{5}{12}\left(\frac{6}{11}+\frac{5}{11}\right)\right]+\frac{7}{12}\)
\(=\frac{5}{12}+\frac{7}{12}\)
\(=\frac{12}{12}=1\)
d) \(-\frac{5}{9}+\frac{14}{9}\left(\frac{3}{4}-\frac{2}{5}\right):49\)
\(=-\frac{5}{9}+\frac{14}{9}\left(\frac{15}{20}-\frac{8}{20}\right):49\)
\(=-\frac{5}{9}+\frac{14}{9}.\frac{7}{20}.\frac{1}{49}\)
\(=-\frac{5}{9}+\frac{7}{9}.\frac{7}{10}.\frac{1}{7.7}\)
\(=-\frac{5}{9}+\frac{1}{90}\)
\(=-\frac{50}{90}+\frac{1}{90}=-\frac{49}{90}\)
\(a)15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)
\(=15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\)
\(=15\frac{3}{13}-8\frac{3}{13}-3\frac{4}{7}\)
\(=7-3\frac{4}{7}\)
\(=6\frac{7}{7}-3\frac{4}{7}=3\frac{3}{7}\)
\(b)\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=7\frac{4}{9}-4\frac{7}{11}-3\frac{4}{9}\)
\(=7\frac{4}{9}-3\frac{4}{9}-4\frac{7}{11}\)
\(=4-4\frac{7}{11}=\frac{7}{11}\)
a)\(\frac{-10}{13}+\frac{8}{17}-\frac{3}{13}+\frac{12}{17}-\frac{11}{20}\)
= \(\frac{-10}{13}+\frac{8}{17}+\frac{-3}{13}+\frac{12}{17}+\frac{-11}{20}\)
=\(\left(\frac{-10}{13}+\frac{-3}{13}\right)+\left(\frac{8}{17}+\frac{12}{17}\right)+\frac{-11}{20}\)
=\(\frac{-13}{13}+\frac{20}{17}+\frac{-11}{20}\)
= \(\frac{-127}{340}\)
b) \(\frac{3}{4}+\frac{-5}{6}-\frac{11}{-12}\)
= \(\frac{3}{4}+\frac{-5}{6}+\frac{11}{12}\)
= \(\frac{9}{12}+\frac{-10}{12}+\frac{11}{12}\)
=\(\frac{10}{12}=\frac{5}{6}\)
c) \(\left[13.\frac{4}{9}+2.\frac{1}{9}\right]-3.\frac{4}{9}\)
= \(13+2.\left(\frac{4}{9}+\frac{1}{9}\right)-3.\frac{4}{9}\)
=\(15.\frac{5}{9}-3.\frac{4}{9}\)
=\(\left[15-3.\left(\frac{5}{9}-\frac{4}{9}\right)\right]\)
=\(12.\frac{1}{9}\)
=\(\frac{4}{3}\)
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