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T
5 tháng 5 2019
#)Trả lời :
a) \(\frac{-1}{2}+\frac{2}{7}-\frac{5}{28}\)
\(=\frac{-3}{14}-\frac{5}{28}\)
\(=\frac{-11}{28}\)
b) \(5\frac{3}{4}+\left(\frac{1}{9}-1\frac{3}{4}\right)\)
\(=\left(5\frac{3}{4}-1\frac{3}{4}\right)+\frac{1}{9}\)
\(=4+\frac{1}{9}\)
\(=4\frac{1}{9}=\frac{37}{9}\)
c) \(\frac{-5}{7}x\frac{2}{11}+\frac{-5}{7}x\frac{9}{11}+3\frac{5}{7}\)
\(=\frac{-5}{7}x\left(\frac{2}{11}+\frac{9}{11}\right)+3\frac{5}{7}\)
\(=\frac{-5}{7}x1+3\frac{5}{7}\)
\(=\frac{-5}{7}+3\frac{5}{7}\)
\(=\frac{-5}{7}+\frac{26}{7}\)
\(=\frac{21}{7}=3\)
#~Will~be~Pens~#
24 tháng 4 2019
a,\(\frac{7}{10}\cdot\frac{4}{9}+\frac{3}{10}\cdot\frac{4}{9}-1\frac{7}{9}\)
\(=\frac{14}{45}+\frac{2}{15}-\frac{16}{9}\)
\(=\frac{14}{45}+\frac{6}{45}-\frac{80}{45}\)
\(=\frac{-60}{45}=\frac{-4}{3}\)
b,\(\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{5}{4}-\frac{2}{3}\right)\cdot\left(-3\right)^2+\frac{5}{9}\cdot30\%\)
\(=\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{7}{12}\right)\cdot9+\frac{5}{9}\cdot\frac{3}{10}\)
\(=\frac{-5}{6}+\frac{7}{3}+\frac{1}{6}\)
\(=\frac{-5}{6}+\frac{14}{6}+\frac{1}{6}\)
=\(=\frac{10}{6}=\frac{5}{3}\)
11 tháng 8 2016
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
11 tháng 8 2016
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé
2 tháng 8 2018
1) \(-1,6:\left(1+\frac{8}{12}\right)\)
\(=-1,6:\left(1+\frac{2}{3}\right)\)
\(=-1,6:1-1,6:\frac{2}{3}\)
\(=-1,6+\frac{-1,6.3}{2}\)
\(=-1,6-2,4\)
\(=4\)
2) \(\frac{3}{8}-\frac{4}{5}-\frac{-17}{40}\)
\(=\frac{15-32+17}{40}\)
=0
3) \(\frac{3}{4}-\frac{16}{32}+\frac{4}{-3}\)
\(=\frac{3}{4}-\frac{2}{4}+\frac{-4}{3}\)
\(=\frac{9-6-16}{12}\)
\(=\frac{-13}{12}\)
4) \(\frac{-4}{7}+\frac{2}{3}.\frac{-9}{14}\)
\(=\frac{-4}{7}+\frac{-3}{7}=-1\)
5) \(2\frac{3}{4}-\left(\frac{1}{7}+1\frac{3}{4}\right)\)
\(=\frac{11}{4}-\left(\frac{1}{7}+\frac{7}{4}\right)\)
\(=\frac{11}{4}-\frac{4+49}{28}\)
\(=\frac{11}{4}-\frac{53}{28}\)
\(=\frac{77-53}{28}=\frac{24}{28}=\frac{6}{7}\)
6) \(8\frac{2}{7}-\left(3\frac{4}{9}+4\frac{2}{7}\right)\)
\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)=\frac{522-217+270}{63}\)
\(=\frac{575}{63}\)
7 tháng 5 2019
\(\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}=\frac{\frac{2.14-33+5.12}{132}}{\frac{5.11+312-84}{132}}=\frac{2.14-33+5.12}{5.11+312-84}=\frac{115}{103}\)
\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.9}+...+\frac{2}{196.199}=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.9}+...+\frac{3}{196.199}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{196}-\frac{1}{199}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{199}\right)\)
\(=\frac{2}{3}.\frac{198}{199}\)
\(=\frac{132}{199}\)