\(a,MSC:180\\ \dfrac{17}{12}=\dfrac{17.15}{12.15}=\dfrac{255}{180};\dfrac{31}{18}=\dfrac{31.10}{18.10}=\dfrac{310}{180};\dfrac{8}{15}=\dfrac{8.12}{15.12}=\dfrac{96}{180}\\ b,MSC:75\\ \dfrac{7}{15}=\dfrac{7.5}{15.5}=\dfrac{35}{75};\dfrac{8}{25}=\dfrac{8.3}{25.3}=\dfrac{24}{75};\dfrac{11}{75}=\dfrac{11}{75}\)

Bài 2:

a: -2*(-27)=54

6*9=54

=>Hai phân số này bằng nhau

b: -1/-5=1/5=5/25<>4/25

Bài 3:

a: =>16/x=-4/5

=>x=-20

b: =>(x+7)/15=-2/3

=>x+7=-10

=>x=-17

a) \(\dfrac{-2}{9}\) và \(\dfrac{6}{-27}\)

\(\dfrac{6}{-27}=\dfrac{6:\left(-3\right)}{\left(-27\right):\left(-3\right)}=\dfrac{-2}{9}\)

Vậy \(\dfrac{-2}{9}=\dfrac{6}{-27}\)

b) \(\dfrac{-1}{-5}\) và \(\dfrac{4}{25}\)

\(\dfrac{-1}{-5}=\dfrac{\left(-1\right).\left(-5\right)}{\left(-5\right).\left(-5\right)}=\dfrac{5}{25}\)

Do \(5\ne4\Rightarrow\dfrac{5}{25}\ne\dfrac{4}{25}\)

Vậy \(\dfrac{-1}{-5}\ne\dfrac{4}{25}\)

Bài 3

a) \(\dfrac{-28}{35}=\dfrac{16}{x}\)

\(x=\dfrac{35.16}{-28}\)

\(x=-20\)

b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}\)

\(\left(x+7\right).36=15.\left(-24\right)\)

\(36x+252=-360\)

\(36x=-360-252\)

\(36x=-612\)

\(x=\dfrac{-612}{36}\)

\(x=-17\)

a) \(\dfrac{20}{12}\)          b) \(\dfrac{15}{15}\)          c) \(\dfrac{24}{42}\)

d) \(\dfrac{26}{39}\)          e) \(\dfrac{0}{8}\)          g) \(\dfrac{15}{3}\)

a) \(\dfrac{15}{3}\)

b)\(\dfrac{15}{15}\)

c) \(\dfrac{24}{42}\)

d) \(\dfrac{26}{39}\)

e)\(\dfrac{0}{8}\)

f) \(\dfrac{15}{15}\)

2/

a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)

\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)

-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)

b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)

-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)

c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)

\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)

-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)

a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)

c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)

a) 

b) 

c) \(\dfrac{25}{100}=\dfrac{10}{40}\)

 = \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)

 = \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)

 = \(\dfrac{5}{2}.\dfrac{100}{101}\)

 = \(\dfrac{250}{101}\)

a: \(\dfrac{5}{7}=\dfrac{5\cdot11}{7\cdot11}=\dfrac{55}{77}\)

\(\dfrac{9}{11}=\dfrac{9\cdot7}{11\cdot7}=\dfrac{63}{77}\)

b: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot9}{7\cdot9}=\dfrac{54}{63}\)

\(-\dfrac{12}{54}=\dfrac{-2}{9}=\dfrac{-2\cdot7}{9\cdot7}=-\dfrac{14}{63}\)

c: \(\dfrac{-11}{30}=\dfrac{-11\cdot4}{30\cdot4}=\dfrac{-44}{120}\)

\(\dfrac{-17}{-40}=\dfrac{17}{40}=\dfrac{17\cdot3}{40\cdot3}=\dfrac{51}{120}\)

d: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot3}{7\cdot3}=\dfrac{18}{21}\)

\(\dfrac{-12}{36}=\dfrac{-1}{3}=\dfrac{-1\cdot7}{3\cdot7}=\dfrac{-7}{21}\)

A = \(\dfrac{3}{2}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{12}\) - \(\dfrac{9}{20}\) + \(\dfrac{11}{30}\) - \(\dfrac{13}{42}\) + \(\dfrac{15}{56}\) - \(\dfrac{17}{72}\)

A = (1 + \(\dfrac{1}{2}\)) - (\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)) - (\(\dfrac{1}{4}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\) + \(\dfrac{1}{6}\)) - (\(\dfrac{1}{6}\) + \(\dfrac{1}{7}\)) + (\(\dfrac{1}{7}\) + \(\dfrac{1}{8}\)) - (\(\dfrac{1}{8}\) + \(\dfrac{1}{9}\))

A = 1 + \(\dfrac{1}{2}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\)

A = 1 - \(\dfrac{1}{9}\)

\(A=\left(1+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{7}+\dfrac{1}{8}\right)-\left(\dfrac{1}{8}+\dfrac{1}{9}\right)\)

\(A=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{8}-\dfrac{1}{9}\)

\(A=1+\dfrac{1}{9}=\dfrac{10}{9}\)

a: \(\dfrac{11}{42}=\dfrac{11\cdot1}{42\cdot1}=\dfrac{11}{42}\)

\(\dfrac{5}{6}=\dfrac{5\cdot7}{6\cdot7}=\dfrac{35}{42}\)

 b: \(\dfrac{40}{7}=\dfrac{40\cdot4}{7\cdot4}=\dfrac{160}{28}\)

\(\dfrac{10}{28}=\dfrac{10\cdot1}{28\cdot1}=\dfrac{10}{28}\)

c: \(\dfrac{4}{15}=\dfrac{4\cdot4}{15\cdot4}=\dfrac{16}{60}\)

\(\dfrac{14}{60}=\dfrac{14\cdot1}{60\cdot1}=\dfrac{14}{60}\)

a)\(\dfrac{11}{42}=\dfrac{35}{42}\)

b) \(\dfrac{160}{28}=\dfrac{10}{28}\)

c) \(\dfrac{16}{60}=\dfrac{14}{60}\)