\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)

\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)

\(A=\frac{2\left(10-1\right)}{5-14}\)

\(A=\frac{2.9}{-9}\)

\(A=-2\)

Vậy \(A=-2\)

\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)

\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)

\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)

\(B=81.\frac{18}{5}.\frac{2}{9}\)

\(B=\frac{324}{5}\)

Vậy \(B=\frac{324}{5}\)

Chúc bạn học tốt ~ ( mỏi tay qué >_< ) 

M=3/1.2-5/2.3+7/3.4-9/4.5+11/5.6-13/6.7+15/7.8+17/8.9

   =(1/1.1+2/1.2)-(2/2.3+3/2.3)+(3/3.4+4/3.4)-(4/4.5+5/4.5)+...+(8/8.9+9/8.9)(phần ... là làm tương tự nhé)

   =1/2+1-(1/3+1/2)+(1/4+1/3)-(1/5+1/4)+...+(1/9+1/8)(phần ... là làm tương tự nhé)

   =1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/8-1/8)-1/9

   =1+0+0+0+...+0-1/9

   =1-1/9

   =8/9

I'm sorry, I can't help you

\(\frac{5}{9}-\frac{7}{13}-\frac{5}{9}.\frac{3}{13}+\frac{-5}{9}.\frac{6}{9}\)

= \(\frac{5}{9}-\frac{7}{13}-\frac{5}{3}.\frac{1}{13}+\frac{-5}{3}.\frac{2}{9}\)

=\(\frac{5}{9}-\frac{7}{13}-\frac{5}{39}+\frac{-10}{27}\)

=\(\frac{5}{9}+\frac{-7}{13}+\frac{-5}{39}+\frac{-10}{27}\)

=\(\frac{-13}{27}\)

Chúc bạn học tốt nhea ♥♥♥

\(\frac{5}{9}-\frac{7}{13}-\frac{5}{9}.\frac{3}{13}+\frac{-5}{9}.\frac{61}{9}\)

\(=\frac{5}{9}\left(1-\frac{3}{13}\right)-\frac{7}{13}-\frac{5}{9}.\frac{61}{9}\)

\(=\frac{5}{9}.\frac{10}{13}-\frac{5}{9}.\frac{61}{9}-\frac{7}{13}\)

\(=\frac{5}{9}\left(\frac{10}{13}-\frac{61}{9}\right)-\frac{7}{13}\)\(=-\frac{314}{81}\)

chịu thôi

B=64,8

Mình làm như thế này nek

\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)

\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)

\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)

\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

dài quá! mình bó tay luôn!

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

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

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

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

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

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

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

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

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

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