a: Ta có: \(\left(-\frac57-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\left(3-\frac34\right)\)

\(=\left(-\frac{6435}{9009}-\frac{7007}{9009}+\frac{7371}{9009}-\frac{7623}{9009}\right)\cdot\frac94\)

\(=-\frac{13694}{9009}\cdot\frac94=\frac{-6847}{2}\cdot\frac{1}{1001}=\frac{-6847}{2002}\)

b: \(\left(\frac{10}{21}+\frac{14}{27}-\frac{6}{11}+\frac{22}{39}\right):\left(2-\frac23\right)\)

\(=\left(\frac{12870}{27027}+\frac{14014}{27027}-\frac{14742}{27027}+\frac{15246}{27027}\right):\frac43\)

\(=\frac{27388}{27027}\cdot\frac34=\frac{6847}{9009}\)

cho mik xin bài giải với ạ

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

1) \(125^5:25^7\)

\(=\left(5^3\right)^5:\left(5^2\right)^7\)

\(=5^{15}:5^{14}\)

= 5

2) \(27^8:9^9\)

\(=\left(3^3\right)^8:\left(3^2\right)^9\)

\(=3^{24}:3^{18}\)

\(=3^6\)

3) \(36^5:6^8\)

\(=\left(6^2\right)^5:6^8\)

\(=6^{10}:6^8\)

\(=6^2\)

4) \(49^6:7^{10}\)

\(=\left(7^2\right)^6:7^{10}\)

\(=7^{12}:7^{10}=7^2\)

5) \(7^{20}:49^9\)

\(=7^{20}:\left(7^2\right)^9\)

\(=7^{20}:7^{18}=7^2\)

6) \(\frac{1}{2^{10}}:\frac{1}{8^3}\)

\(=\frac{1}{2^{10}}:\frac{1}{\left(2^3\right)^3}\)

\(=\frac{1}{2^{10}}:\frac{1}{2^9}=\frac{1}{2^{10}}.\frac{2^9}{1}=\frac{1}{2}\)

7) \(\left(-\frac{1}{2}\right)^{21}:\frac{1}{4^{10}}\)

\(=\frac{\left(-1\right)^{21}}{2^{21}}:\frac{1}{\left(2^2\right)^{10}}\)

\(=-\frac{1}{2^{21}}:\frac{1}{2^{20}}=-\frac{1}{2^{21}}.\frac{2^{20}}{1}\)

\(=-\frac{1}{2}\)

8) \(\frac{1}{16^5}:\left(-\frac{1}{2}\right)^{18}\)

\(=\frac{1}{\left(2^4\right)^5}:\frac{\left(-1\right)^{18}}{2^{18}}\)

\(=\frac{1}{2^{20}}:\frac{1}{2^{18}}\)

\(=\frac{1}{2^{20}}.\frac{2^{18}}{1}=\frac{1}{4}\)

9) \(\frac{1}{5^{30}}:\frac{1}{25^{14}}\)

\(=\frac{1}{5^{30}}:\frac{1}{\left(5^2\right)^{14}}\)

\(=\frac{1}{5^{30}}:\frac{1}{5^{28}}=\frac{1}{25}\)

Bài 1:

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)

A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)

A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))

A = 1 - 1 - \(\frac{228}{210}\)

A = 0 - \(\frac{128}{210}\)

A = - \(\frac{64}{105}\)

Bài 2:

B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)

B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)

B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))

B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))

B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))

B = 3 - \(\frac{97}{65}\)

B = \(\frac{195}{65}\) - \(\frac{97}{65}\)

B = \(\frac{98}{65}\)