\(\frac{3}{5}.\frac{13}{46}-\frac{1}{10}.\frac{16}{23}\)

= \(\frac{39}{230}-\frac{8}{115}\)

= \(\frac{1}{10}\)

\(\frac{1}{10}\) nha bạn

\(\frac{1}{10}\) nha bạn

\(\begin{array}{l}B = \left( {\frac{{ - 3}}{{13}}} \right) + \frac{{16}}{{23}} + \left( {\frac{{ - 10}}{{13}}} \right) + \frac{5}{{11}} + \frac{7}{{23}}\\ = \left[ {\left( {\frac{{ - 3}}{{13}}} \right) + \left( {\frac{{ - 10}}{{13}}} \right)} \right] + \left[ {\frac{{16}}{{23}} + \frac{7}{{23}}} \right] + \frac{5}{{11}}\\ =  - 1 + 1 + \frac{5}{{11}}\\ = \frac{5}{{11}}\end{array}\)

`B= ( (-3)/13 + (-10)/13) + (16/23 + 7/23 ) +5/11`

`B= -13/13 + 23/23 +5/11`

`B=-1+1+5/11`

`B=0+5/11`

`B=5/11`

\(B=\left(-\dfrac{3}{13}\right)+\dfrac{16}{23}+\left(-\dfrac{10}{13}\right)+\dfrac{5}{11}+\dfrac{7}{23}\)

\(B=\left[\left(-\dfrac{3}{13}\right)+\left(-\dfrac{10}{13}\right)\right]+\left(\dfrac{16}{23}+\dfrac{7}{23}\right)+\dfrac{5}{11}\)

\(B=\left(-1\right)+1+\dfrac{5}{11}\)

\(B=\dfrac{5}{11}\)

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=\(\frac{-41000}{973}\)

giup mk vs cần gấp nk

\(\frac{\left(13\frac{1}{4}-2\frac{5}{7}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}=\frac{\left(\frac{53}{4}-\frac{19}{7}-\frac{65}{6}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{13}{10}+\frac{10}{3}\right):\left(\frac{37}{3}-\frac{100}{7}\right)}\)

\(=\frac{\left(\frac{1113}{84}-\frac{228}{84}-\frac{910}{84}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{39}{30}+\frac{100}{30}\right):\left(\frac{259}{21}-\frac{300}{21}\right)}\)

\(=\frac{\frac{-25}{84}.\frac{5751}{25}+\frac{187}{4}}{\frac{139}{30}:\frac{-41}{21}}\)

\(=\frac{\frac{-1917}{28}+\frac{1309}{28}}{\frac{139}{30}.\frac{-21}{41}}\)

\(=\frac{\frac{-608}{28}}{\frac{-973}{410}}=\frac{-152}{7}.\frac{410}{-973}=\frac{62320}{6811}\)

a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)

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

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

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

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

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

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

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

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

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

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