do dịch covid mình quên dạng này rồi ://

hi bro

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

A<B

A < B

Sửa đề : \(-\dfrac{19}{37}=\dfrac{19}{37}\)

\(\dfrac{19}{37}-\dfrac{31}{24}-\dfrac{19}{37}+1\dfrac{7}{24}-\dfrac{2}{13}\)

\(=\dfrac{19}{37}-\dfrac{31}{24}+\dfrac{19}{37}+\dfrac{31}{24}-\dfrac{2}{13}\)

\(=\left(\dfrac{19}{37}-\dfrac{19}{37}\right)+\left(-\dfrac{31}{24}+\dfrac{31}{24}\right)-\dfrac{2}{13}=-\dfrac{2}{13}\)

ĐỀ ĐÚNG MÀ A

1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)

\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)

\(=\dfrac{-1}{2}+\dfrac{4}{5}\)

\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)

2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)

\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)

\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)

3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)

\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)

\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{17}{7}\)

4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)

\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)

\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)

\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)

a) x vô nghĩa

b) x=0,8;x=-0,8

bn giải ra đi

A = \(1\dfrac{13}{15}.25.\dfrac{1}{100}.3+\left(\dfrac{8}{15}-\dfrac{78}{60}\right)\div1\dfrac{23}{24}\)

A = \(\dfrac{1.15+13}{15}.25.\dfrac{1}{100}.3+\left(\dfrac{32}{60}-\dfrac{78}{60}\right)\div\dfrac{1.24+23}{24}\)

A = \(\dfrac{28}{15}.25.\dfrac{1}{100}.3+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)

A = \(\left(\dfrac{28}{15}.25\right).\left(\dfrac{1}{100}.3\right)+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)

A = \(\left(\dfrac{28.25}{15}\right).\left(\dfrac{1.3}{100}\right)+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)

A = \(\dfrac{140}{3}.\dfrac{3}{100}+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)

A = \(\dfrac{7}{5}+\left(-\dfrac{92}{235}\right)=\dfrac{237}{235}\)

\(A=1\dfrac{13}{15}.25.\dfrac{1}{100}.3+\left(\dfrac{8}{15}-\dfrac{78}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.25.\dfrac{1}{100}.3-\dfrac{23}{30}:\dfrac{47}{24}\\ =\dfrac{7}{5}-\dfrac{23}{30}.\dfrac{24}{47}\\ =\dfrac{7}{5}-\dfrac{92}{235}\\ =\dfrac{7.47}{235}-\dfrac{92}{235}\\ =\dfrac{237}{235}=1\dfrac{2}{235}\)

a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)

=>2/5x=8/5

=>x=4

b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)

\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)

=>1/3x=-6

=>x=-18

c: =>2|x-1/3|=0,24-4/5=-0,56<0