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9 tháng 2 2017
1. \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
= \(\frac{5}{9}\) .(\(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\) )
= \(\frac{5}{9}\) . 1 = \(\frac{5}{9}\)
25 tháng 4 2017
\(-\frac{3}{5}.\frac{5}{7}+-\frac{3}{5}.\frac{3}{7}+-\frac{3}{5}.\frac{6}{7}=-\frac{3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)=-\frac{3}{5}.2=-\frac{6}{5}\)
\(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{4}{3}=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{4}{3}=\frac{1}{3}.2-\frac{4}{3}=\frac{2}{3}-\frac{4}{3}=-\frac{2}{3}\)
\(\frac{4}{19}.\frac{-3}{7}+-\frac{3}{7}.\frac{15}{19}+\frac{5}{7}=-\frac{3}{7}\left(\frac{4}{19}+\frac{15}{19}\right)+\frac{5}{7}=-\frac{3}{7}+\frac{5}{7}=\frac{2}{7}\)
\(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\)
25 tháng 4 2017
\(a,6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)=6\frac{4}{5}-1\frac{2}{3}-3\frac{4}{5}=6\frac{4}{5}-3\frac{4}{5}-1\frac{2}{3}=3-1\frac{2}{3}=\frac{4}{3}\)
\(b,6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)=6\frac{5}{7}-2\frac{5}{7}-1\frac{3}{4}=\frac{9}{4}\)
\(c,7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)=7\frac{5}{9}-3\frac{5}{9}-2\frac{3}{4}=4-2\frac{3}{4}=\frac{5}{4}\)
mk nghĩ là phần d như thế này cơ \(7\frac{5}{11}\left(2\frac{3}{7}+3\frac{5}{11}\right)\)
\(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)=7\frac{5}{11}-3\frac{5}{11}-2\frac{3}{7}=4-2\frac{3}{7}=\frac{11}{7}\)
1 tháng 8 2020
a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)
\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)
\(=4-\frac{4}{7}=\frac{24}{7}\)
b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)
\(=8+\frac{7}{11}=\frac{95}{11}\)
c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)
\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)
\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)
\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)
d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)
\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)
\(=5\cdot\frac{7}{35}=1\)
e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)
\(=\frac{42}{43}\)
3 tháng 5 2020
Bài 1:
a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)
\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)
\(=\frac{-3}{5}\)
b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)
\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)
\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)
c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)
\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)
\(=3+2=5\)
d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)
\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)
\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)
e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)
\(=-1+1+\frac{-7}{9}\)
\(=-\frac{7}{9}\)
f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)
\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)
\(=1+\frac{-1}{7}=\frac{7}{7}+\frac{-1}{7}=\frac{6}{7}\)
18 tháng 4 2019
\(a)\frac{-3}{4}+\frac{3}{7}+-\frac{1}{4}+\frac{4}{9}+\frac{4}{7}\)
\(=\left(\frac{-3}{4}+-\frac{1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)
\(=-1+1+\frac{4}{9}\)
\(=\frac{4}{9}\)
18 tháng 4 2019
\(\frac{-3}{4}+\frac{3}{7}+\frac{-1}{4}+\frac{4}{9}+\frac{4}{7}\)
\(=\left(\frac{-3}{4}+\frac{-1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)
\(=\left(-1\right)+1+\frac{4}{9}\)
\(=0+\frac{4}{9}\)
\(=\frac{4}{9}\)
\(F=\frac{5}{9}+\frac{4}{9}\cdot\frac{3}{7}+\frac{4}{9}\cdot\frac{4}{7}\)
\(F=\frac{5}{9}+\frac{4}{9}\cdot\left(\frac{3}{7}+\frac{4}{7}\right)\)
\(F=\frac{5}{9}+\frac{4}{9}\cdot1\)
'\(F=\frac{5}{9}+\frac{4}{9}\)
\(F=1\)