a)

\(\begin{array}{l}\frac{3}{7}.\left( { - \frac{1}{9}} \right) + \frac{3}{7}.\left( { - \frac{2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} + \frac{-2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} - \frac{6}{9}} \right)\\ = \frac{3}{7}.\frac{{ - 7}}{9} = \frac{{ - 1}}{3}\end{array}\)                 

b)

\(\begin{array}{l}\left( {\frac{{ - 7}}{{13}}} \right).\frac{5}{{12}} + \left( {\frac{{ - 7}}{{13}}} \right).\frac{7}{{12}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.\left( {\frac{5}{{12}} + \frac{7}{{12}}} \right) + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.1 + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 13}}{{13}}\\ = -1\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right]:\frac{5}{9} + \left( {\frac{4}{7} - \frac{1}{3}} \right):\frac{5}{9}\\ = \left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right].\frac{9}{5} + \left( {\frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left( {\frac{{ - 2}}{3} + \frac{3}{7} + \frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left[ {\left( {\frac{{ - 2}}{3} - \frac{1}{3}} \right) + \left( {\frac{3}{7} + \frac{4}{7}} \right)} \right].\frac{9}{5}\\ = \left( { - 1 + 1} \right).\frac{9}{5}\\ = 0.\frac{9}{5} = 0\end{array}\)

d)

\(\begin{array}{l}\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{2}{3}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{{10}}{{15}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 9}}{15}\\= \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 3}}{5}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{5}{9}.\frac{{ - 5}}{3}\\ = \frac{5}{9}.\left( {\frac{{ - 22}}{3} - \frac{5}{3}} \right)\\ = \frac{5}{9}.\frac{-27}{3}= \frac{5}{9}.\left( { - 9} \right) =  - 5\end{array}\)

e)

\(\begin{array}{l}\frac{3}{5} + \frac{3}{{11}} - \left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{{ - 2}}{{97}}} \right) - \frac{1}{{35}} - \frac{3}{4} + \left( {\frac{{ - 23}}{{44}}} \right)\\ = \frac{3}{5} + \frac{3}{{11}} + \frac{3}{7} - \frac{2}{{97}} - \frac{1}{{35}} - \frac{3}{4} - \frac{{23}}{{44}}\\ = \left( {\frac{3}{5} + \frac{3}{7} - \frac{1}{{35}}} \right) + \left( {\frac{3}{{11}} - \frac{3}{4} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \left( {\frac{{21}}{{35}} + \frac{{15}}{{35}} - \frac{1}{{35}}} \right) + \left( {\frac{{12}}{{44}} - \frac{{33}}{{44}} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \frac{35}{{35}}+ \frac{-44}{{44}}- \frac{2}{{97}}\\= 1 + \left( { - 1} \right) - \frac{2}{{97}}\\ =  - \frac{2}{{97}}\end{array}\)

\(a)\frac{2}{{15}} + \left( {\frac{{ - 5}}{{24}}} \right) = \frac{{16}}{{120}} + \left( {\frac{{ - 25}}{{120}}} \right) = \frac{{ - 9}}{{120}} = \frac{{ - 3}}{{40}}\)          

 b) \(\left( {\frac{{ - 5}}{9}} \right) - \left( { - \frac{7}{{27}}} \right) = \left( {\frac{{ - 15}}{{27}}} \right) + \frac{7}{{27}} = \frac{{ - 8}}{{27}}\)          

 c)\(\left( { - \frac{7}{{12}}} \right) + 0,75 = \left( { - \frac{7}{{12}}} \right) + \frac{75}{100} \\= \left( { - \frac{7}{{12}}} \right) + \frac{3}{4} \\= \left( { - \frac{7}{{12}}} \right) + \frac{9}{{12}} = \frac{2}{{12}} = \frac{1}{6}\)

d)\(\left( {\frac{{ - 5}}{9}} \right) - 1,25 =\left( {\frac{{ - 5}}{9}} \right) - \frac{125}{100} = \left( {\frac{{ - 5}}{9}} \right) - \frac{5}{4}\\ = \left( {\frac{{ - 20}}{{36}}} \right) - \frac{{45}}{{36}} = \frac{{ - 65}}{{36}}\)       

 e)\(0,34.\frac{{ - 5}}{{17}} =\frac{{34}}{{100}}.\frac{{ - 5}}{{17}} = \frac{{17}}{{50}}.\frac{{ - 5}}{{17}} = \frac{{ - 1}}{{10}}\)                       

g) \(\frac{4}{9}:\left( { - \frac{8}{{15}}} \right) = \frac{4}{9}.\left( { - \frac{{15}}{8}} \right) = \frac{{ - 5}}{6}\)

h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right) = \frac{5}{3}:\frac{5}{2} = \frac{5}{3}.\frac{2}{5} = \frac{2}{3}\)       

i) \(\frac{2}{5}.\left( { - 1,25} \right) = \frac{2}{5}.\frac{{ - 125}}{100} = \frac{2}{5}.\frac{{ - 5}}{4} = \frac{{ - 1}}{2}\)                     

k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9} = \left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).\frac{{28}}{9}\\ = \frac{{ - 3.3.5.7.4}}{{5.\left( { - 7} \right).3.3}} = 4\)

b)-5/12 . 2/11+(-5/12) . 9/11+5/12

=(-5/12+5/12).(2/11+9/11)

=0.1=0

c)6/9/10+(2/5-1/10)

=69/10+3/10

=36/5

\(b)\)\(\frac{-5}{12}.\frac{2}{11}+\frac{-5}{12}.\frac{9}{11}+\frac{5}{12}\)

\(=\frac{-5}{12}.\frac{2}{11}+\frac{-5}{12}.\frac{9}{11}+\frac{5}{12}.1\)

\(=\frac{-5}{12}.\left(\frac{2}{11}+\frac{9}{11}+1\right)\)

\(=\frac{-5}{12}.2\)

\(=\frac{-5}{6}\)

\(c)\)\(6\frac{9}{10}+\left(\frac{2}{5}-\frac{1}{10}\right)\)

\(=6\frac{9}{10}+\frac{2}{5}-\frac{1}{10}\)

\(=\left(6\frac{9}{10}-\frac{1}{10}\right)+\frac{2}{5}\)

\(=6\frac{4}{5}+\frac{2}{5}\)

\(=6\frac{6}{5}=7\frac{1}{5}\)

chịu thôi

B=64,8

a,

A = 5.(2² . 3² )⁹. (2²)⁶ - 2. (2².3)¹⁴.3⁴ / 5.2²⁸.3¹⁸ - 7.2²⁹.3¹⁸

A = 5.2¹⁸.3¹⁸.2¹² - 2. 2^28. 3¹⁴. 3⁴ / 5. 2²⁸. 3¹⁸ - 7. 2²⁹. 3¹⁸

A = 5.2³⁰.3¹⁸ - 2²⁹. 3¹⁸ /  5.2 ²⁸.3¹⁸ - 7.2²⁹. 3¹⁸

A = 2²⁹. 3¹⁸(5.2-1) / 2²⁸. 3¹⁸ (5 -7.2)

A = 2.9 / -9 = -2

Vậy A = -2

b,

B =81.( 12 - 12/7- 12/289 - 12/85 /  4 - 4/7 - 4/289 - 4/85 :

5 + 5/13 + 5/169 + 5/91 /  6 + 6/13 + 6/169 + 5/91 ) x 158158158

B = 81.[ 12.( 1 - 1/7 - 1/289 - 1/85) / 4.( 1 - 1/7 - 1/289 - 1/85 :

5.(1 + 1/13 + 1/169 + 1/91) / 6.(1+ 1/13 +1/169 + 1/91)] . 2/9

B = 81.[ 3 : 5/6 ] .2/9

B = 81.18/5 . 2/9

B = 1458/5 . 2/9

B = 324/5.

Vậy B = 324/5

CHÚC HỌC TỐT

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

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

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

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

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

\(A=-2\)

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 >_< ) 

A=\(-\frac{155}{284}\)

B=\(-\frac{43}{3780}\)

A = 0 

B= 3/11

C= -1 

D= -9/10

dễ mà

trừ đi rùi rút gọn

\(C=\left(\frac{3}{5}-\frac{4}{15}\right).\left(\frac{2}{7}-\frac{3}{14}\right)-\left(\frac{5}{9}-\frac{7}{27}\right).\left(1-\frac{3}{5}\right)+\left(1-\frac{11}{12}\right).\left(1+\frac{1}{12}\right)\)

\(C=\left(\frac{9}{15}-\frac{4}{15}\right).\left(\frac{4}{14}-\frac{3}{14}\right)-\left(\frac{15}{27}-\frac{7}{27}\right).\left(\frac{5}{5}-\frac{3}{5}\right)+\left(\frac{12}{12}-\frac{11}{12}\right).\left(\frac{12}{12}+\frac{1}{12}\right)\)

\(C=\frac{5}{15}.\frac{1}{14}.\frac{8}{27}.\frac{2}{5}.\frac{1}{12}.\frac{13}{12}\)

\(C=\frac{5.1.8.2.1.13}{15.14.27.5.12.12}\)

\(C=\frac{5.2.4.2.13}{3.5.14.27.5.4.3.2.2.3}\)

\(C=\frac{13}{3.14.27.5.3.3}\)

\(C=\frac{13}{51030}\)