\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)

\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)

\(A=\frac{5}{13}+\frac{1}{3}=\frac{44}{13}\)

Bạn tham khảo nhé 

Ta có : 

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{5.31-5.\frac{2}{7}-5.\frac{1}{11}+5.\frac{1}{23}}{13.31-13.\frac{2}{7}-13.\frac{1}{11}+13.\frac{1}{23}}+\frac{3.\frac{1}{5}+3.\frac{1}{13}-3.\frac{3}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)

\(A=\frac{5\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}+\frac{3\left(\frac{1}{5}+\frac{1}{13}-\frac{3}{10}\right)}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)

\(A=\frac{5}{13}+\frac{3}{1}=\frac{5}{13}+\frac{39}{13}=\frac{44}{13}\)

Vậy \(A=\frac{44}{13}\)

44/13

Ta có: M=\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}\)+\(\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

=\(\frac{5.\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13.\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}\)+\(\frac{\frac{9}{15}+\frac{9}{39}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}+\frac{3}{10}}\)

=\(\frac{5}{13}\)+\(\frac{9.\left(\frac{1}{15}+\frac{1}{39}-\frac{1}{10}\right)}{3.\left(\frac{1}{39}+\frac{1}{15}-\frac{1}{10}\right)}\)

=\(\frac{5}{13}\)+\(\frac{9}{3}\)

=\(\frac{5}{13}\)+3

=\(\frac{44}{13}\)

bạn kieu cao dương nghich quá các bạn đè bạn ấy xuống đi

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}-\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{155-5\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{403-13\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}-\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{7}{91}+\frac{2}{10}-\frac{3}{10}}\)

\(A=\frac{155-5}{403-13}-\frac{3\left(\frac{1}{5}+\frac{1}{13}\right)-\frac{9}{10}}{\frac{7}{91}+\left(-\frac{1}{10}\right)}\)

\(A=\frac{5}{13}-\frac{\left(-\frac{9}{130}\right)}{\left(-\frac{3}{130}\right)}=\frac{5}{13}-\frac{\frac{9}{130}}{\frac{3}{130}}\)

\(A=\frac{5}{13}-\frac{9}{130}\cdot\frac{130}{3}\)

\(A=\frac{5}{13}-3=-\frac{34}{13}\)

\(B=\frac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)

\(B=\frac{30\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot\left(2\cdot7\right)^5\cdot2^{12}}{54\cdot\left(2\cdot3\right)^{14}\cdot\left(3^2\right)^7-12\cdot\left(2^3\right)^5\cdot7^5}\)

\(B=\frac{30\cdot2^{14}\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{54\cdot2^{14}\cdot3^{14}\cdot3^{14}-12\cdot2^{15}\cdot7^5}\)

\(B=\frac{30\cdot3^{29}-5\cdot2^{17}\cdot7^5}{54\cdot3^{28}-12\cdot2^{15}\cdot7^5}=\frac{30\cdot3-5\cdot2^2}{54-12}=\frac{5}{3}\)

\(A=\frac{\left(140\frac{7}{30}-138\frac{5}{12}\right):18\frac{1}{6}}{0,002}\)

\(A=\frac{\left(\frac{4207}{30}-\frac{1661}{12}\right):\frac{109}{6}}{\frac{1}{500}}\)

\(A=\frac{\left(\frac{4207}{30}-\frac{1661}{12}\right)\times\frac{6}{109}}{\frac{1}{500}}\)

\(A=\left(\frac{4207}{30}-\frac{1661}{12}\right)\times\frac{6}{109}\times500\)

\(A=\frac{109}{60}\times\frac{6}{109}\times500\)

\(A=\frac{1}{10}\times500\)

\(A=50\)

\(B=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}\)

\(B=\frac{5\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}\)

\(B=\frac{5}{13}\)

a, 1 - 7x = 3x - 4

=> -7x - 3x = - 4 - 1

=> - 10x = - 5

=> x = 1/2

vậy_

b, đặt  \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3A-A=1-\frac{1}{3^{99}}\)

\(A=\frac{1-\frac{1}{3^{99}}}{2}\)

mk chỉ bt lm mấy phần hui à!

d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)

\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)

e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)

mk xl,mk ko bt lm các phần còn lại!

a)

\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}.\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.\frac{23}{23}\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)                                   

b)

\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ =\frac{{17}}{{13}}. (\frac{-9}{9}+\frac{5}{5})\\= \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)          

d)

\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ =  - 2\end{array}\)