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a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)
b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)
d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)
\(a)\frac{8}{9}x-\frac{2}{3}=\frac{1}{3}x+1\frac{1}{3}\)
\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+1\frac{1}{3}\)
\(\Rightarrow\frac{5}{9}x=\frac{2}{3}+\frac{4}{3}\)
\(\Rightarrow\frac{5}{9}x=2\Rightarrow x=2\div\frac{5}{9}=\frac{18}{5}\)
\(b)(\frac{-2}{5}+\frac{3}{7})-(\frac{4}{9}+\frac{12}{20}-\frac{13}{25})+\frac{7}{35}\)
\(=\frac{1}{35}-(\frac{4}{9}+\frac{3}{5}-\frac{13}{25})+\frac{1}{5}\)
\(=\frac{1}{35}-(\frac{4}{9}+\frac{15}{25}-\frac{13}{25})+\frac{1}{5}\)
\(=\frac{1}{35}-(\frac{4}{9}+\frac{2}{25})+\frac{1}{5}\)
\(=\frac{1}{35}-\frac{118}{25}+\frac{1}{5}\)
Làm nốt
a) \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\)
\(=\frac{15-4}{26}=\frac{11}{26}\)
c) \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}=\frac{-22-1}{23}=\frac{-23}{23}=-1\)
a) 2/7+-3/8+11/7+1/3+1/7+5/-8
=(2/7+11/7+1/7)+(3/8+-5/8)+1/3
=2+2+1/3
=4+1/3
=13/3
b) -3/8+12/25+5/-8+2/-5+13/25
=(-3/8+-5/8)+(12/25+13/25)+-2/5
=-1+1+-2/5
=0+-2/5
=-2/5
c)7/8+1/8*3/8+1/8*5/8
=7/8+1/8*(3/8+5/8)
=7/8+1/8*1
=7/8+1/8
=1
a) 2/7+-3/8+11/7+1/3+1/7+5/-8
=(2/7+11/7+1/7)+(3/8+-5/8)+1/3
=2+2+1/3
=4+1/3
=13/3
b) -3/8+12/25+5/-8+2/-5+13/25
=(-3/8+-5/8)+(12/25+13/25)+-2/5
=-1+1+-2/5
=0+-2/5
=-2/5
c)7/8+1/8*3/8+1/8*5/8
=7/8+1/8*(3/8+5/8)
=7/8+1/8*1
=7/8+1/8
=1
Ta có: \(\left(8-\frac{9}{4}+\frac{2}{7}\right)-\left(-6-\frac{3}{7}+\frac{5}{4}\right)-\left(3+\frac{2}{4}-\frac{9}{7}\right)\)
\(=8-\frac{9}{4}+\frac{2}{7}+6+\frac{3}{7}-\frac{5}{4}-3-\frac{2}{4}+\frac{9}{7}\)
\(=11-4+2\)
=9
1)
\(2.\frac{3}{7}+\left(\frac{2}{9}-1\frac{3}{7}\right)-\frac{5}{3}:\frac{1}{9}\)
\(=\frac{6}{7}+\frac{2}{9}-\frac{10}{7}-\frac{5}{3}.9\)
\(=\left(\frac{6}{7}-\frac{10}{7}\right)+\frac{2}{9}-15\)
\(=\frac{-4}{7}+\frac{2}{9}-\frac{15}{1}\)
\(=\frac{-36}{63}+\frac{14}{63}-\frac{945}{63}\)
\(=\frac{-967}{63}\)
2)
\(-\frac{11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}\)
\(=\frac{-11}{23}\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}\)
\(=\frac{-22}{23}-\frac{1}{23}\)
\(=\frac{-23}{23}\)
\(=-1\)
3)
\(\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{24}\right)\)
\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\left(\frac{4}{24}-\frac{3}{24}-\frac{1}{24}\right)\)
\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\frac{0}{24}\)
\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).0\)
\(=0\)
a. \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\frac{15}{13.2}-\frac{2}{13}=\frac{15}{13.2}-\frac{2.2}{13.2}=\frac{15-4}{26}=\frac{11}{26}\)
C. \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{1}{23}.\left(-11.\frac{6}{7}-11.\frac{8}{7}-1\right)=\frac{1}{23}.\left(-22-1\right)=\frac{1}{23}.\left(-23\right)=-1\)
\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)
\(=\frac{2}{4}=\frac{1}{2}\)
\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)
\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)
\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)
\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)
\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\) \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\) \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\) \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)
\(=\frac{58}{7}-\frac{487}{63}\) \(=\frac{577}{45}-\frac{280}{45}\)
\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\) \(=\frac{33}{5}\)
\(P=M-N\)
\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)
\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)
\(\Rightarrow P=\frac{-272}{45}\)
Vậy P = \(\frac{-272}{45}\)
\(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\)
\(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\)
\(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\)
\(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}=1\)
Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé !
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-\frac{73}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right)=\frac{-91}{23}\)
nha bạn 
^_^