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30 tháng 4 2023

\(-3\dfrac{1}{7}:\left\{\left[2+1\dfrac{3}{5}:\left(\dfrac{4}{3}-1,6\right)\right]+25\%\right\}\)

\(=-\dfrac{22}{7}:\left\{\left[2+\dfrac{8}{5}:\left(\dfrac{4}{3}-\dfrac{8}{5}\right)\right]+\dfrac{1}{4}\right\}\)

\(=-\dfrac{22}{7}:\left[\left(2+\dfrac{8}{5}:\left(-\dfrac{4}{15}\right)\right)+\dfrac{1}{4}\right]\)

\(=-\dfrac{22}{7}:\left(2-6+\dfrac{1}{4}\right)\)

\(=-\dfrac{22}{7}:\left(-\dfrac{15}{4}\right)\)

\(=-\dfrac{22}{7}.\dfrac{-4}{15}\)

\(=\dfrac{88}{105}\)

30 tháng 4 2023

khi nào lên thượng tá đây hả chị

22 tháng 4 2023

1/4 . -3/5 - 7/5 *1/4 +5/2 * 1/5

= -3/20- 7/20 + 10/20

= 0/20

chắc thế

1 tháng 4 2018

\(M=\dfrac{1,6:\left(1\dfrac{3}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(5\dfrac{5}{9}-2\dfrac{1}{4}\right).2\dfrac{1}{4}}+0,6.0,5:\dfrac{2}{5}\)

\(M=\dfrac{1,6:\left(\dfrac{8}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{50}{9}-\dfrac{9}{4}\right).\dfrac{35}{17}}+0,6.0,5:\dfrac{2}{5}\)

\(M=\dfrac{1,6:2}{0,64-\dfrac{1}{25}}+\dfrac{1:\dfrac{4}{7}}{\dfrac{119}{36}.\dfrac{35}{17}}+0,6.0,5:\dfrac{2}{5}\)

\(M=\dfrac{0,8}{0,6}+\dfrac{1,75}{\dfrac{245}{36}}+0,6.0,5:\dfrac{2}{5}\)

\(M=\dfrac{4}{3}+\dfrac{9}{35}+0,6.0,5:\dfrac{2}{5}\)

\(M=\dfrac{167}{105}+0,6.0,5:\dfrac{2}{5}\)

\(M=\dfrac{167}{105}+\dfrac{3}{10}:\dfrac{2}{5}\)

\(M=\dfrac{167}{105}+\dfrac{3}{4}\)

\(M=\dfrac{983}{420}.\)

10 tháng 7 2017

3\(\dfrac{1}{7}\) : \(\left\{\left[2-1\dfrac{3}{5}:\left(\dfrac{4}{3}-1,6\right)\right]-25\%\right\}\)

= \(\dfrac{22}{7}\) : \(\left\{\left[\dfrac{10}{5}-\dfrac{8}{5}:\left(\dfrac{4}{3}-\dfrac{8}{5}\right)\right]-\dfrac{1}{4}\right\}\)

= \(\dfrac{22}{7}\) : \(\left\{\left[\dfrac{10}{5}-\dfrac{8}{5}:\left(\dfrac{20}{15}-\dfrac{24}{15}\right)\right]-\dfrac{1}{4}\right\}\)

= \(\dfrac{22}{7}\) : \(\left\{\left[\dfrac{10}{5}-\dfrac{8}{5}:\dfrac{-4}{15}\right]-\dfrac{1}{4}\right\}\)

= \(\dfrac{22}{7}\) : \(\left\{\left[\dfrac{10}{5}-\dfrac{8}{5}.\dfrac{-15}{4}\right]-\dfrac{1}{4}\right\}\)

= \(\dfrac{22}{7}\) : \(\left\{\left[\dfrac{10}{5}-\dfrac{-30}{5}\right]-\dfrac{1}{4}\right\}\)

= \(\dfrac{22}{7}\) : \(\left\{8-\dfrac{1}{4}\right\}\)

= \(\dfrac{22}{7}\) : \(\left\{\dfrac{32}{4}-\dfrac{1}{4}\right\}\)

= \(\dfrac{22}{7}\) : \(\dfrac{31}{4}\)

= \(\dfrac{22}{7}\) . \(\dfrac{4}{31}\)

= \(\dfrac{88}{217}\)

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11 tháng 5 2017

\(A=\dfrac{1,6:\left(1\dfrac{3}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{5\dfrac{5}{9}-2\dfrac{1}{4}.2\dfrac{2}{17}}+0,6.0,5:\dfrac{2}{5}\)

\(A=\dfrac{\dfrac{8}{5}:\left(\dfrac{8}{5}.\dfrac{5}{4}\right)}{\dfrac{64}{100}-\dfrac{4}{100}}+\dfrac{\left(\dfrac{108}{100}-\dfrac{8}{100}\right):\dfrac{4}{7}}{\left(\dfrac{50}{9}-\dfrac{9}{4}\right).\dfrac{36}{17}}+\dfrac{3.1.5}{5.2.2}\)

\(A=\dfrac{\dfrac{4}{5}}{\dfrac{60}{100}}+\dfrac{1:\dfrac{4}{7}}{\left(\dfrac{200}{36}-\dfrac{81}{36}\right).\dfrac{36}{17}}+\dfrac{3}{4}\)

\(A=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}+\dfrac{\dfrac{7}{4}}{\dfrac{119}{36}.\dfrac{36}{17}}+\dfrac{3}{4}\)

\(A=\dfrac{4}{3}+\dfrac{\dfrac{7}{4}}{7}+\dfrac{3}{4}\)

\(A=\dfrac{16}{12}+\dfrac{3}{12}+\dfrac{9}{12}\)

\(A=\dfrac{28}{12}\) \(=\dfrac{7}{3}\)

a: \(=\dfrac{-7}{8}\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+3+\dfrac{7}{8}=\dfrac{-7}{8}+\dfrac{7}{8}+3=3\)

b: \(=-\dfrac{8}{5}:\dfrac{5}{3}=-\dfrac{24}{25}\)

c: \(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{6}{8}=\dfrac{6}{7}-\dfrac{5}{8}=\dfrac{48}{56}-\dfrac{35}{56}=\dfrac{13}{56}\)

 

2 tháng 4 2017

a) \(\dfrac{-3}{8}-\dfrac{1}{8}:x=-5\dfrac{1}{4}-2\dfrac{1}{4}\)

\(\dfrac{-3}{8}-\dfrac{1}{8}:x=\dfrac{-21}{4}-\dfrac{9}{4}\)

\(\dfrac{-3}{8}-\dfrac{1}{8}:x=\dfrac{-15}{2}\)

\(\dfrac{1}{8}:x=\dfrac{-3}{8}-\dfrac{-15}{2}\)

\(\dfrac{1}{8}:x=\dfrac{57}{8}\)

\(x=\dfrac{1}{8}:\dfrac{57}{8}\)

\(x=\dfrac{1}{57}\)

b) \(\left(1\dfrac{1}{3}-25\%-\dfrac{5}{12}\right)-2x=1,6:\dfrac{3}{5}\)

\(\left(1\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{5}{12}\right)-2x=\dfrac{8}{5}:\dfrac{3}{5}\)

\(\left(\dfrac{4}{3}-\dfrac{1}{4}-\dfrac{5}{12}\right)-2x=\dfrac{8}{3}\)

\(\left(\dfrac{13}{12}-\dfrac{5}{12}\right)-2x=\dfrac{8}{3}\)

\(\dfrac{2}{3}-2x=\dfrac{8}{3}\)

\(2x=\dfrac{2}{3}-\dfrac{8}{3}\)

\(2x=-2\)

\(x=\left(-2\right):2\)

\(x=-1\)

a: =>11(x-3)=6(x-5)

=>11x-33=6x-30

=>5x=3

=>x=3/5

b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3

=>-9/4x+11/12=8/3

=>-9/4x=32/12-11/12=21/12=7/4

=>x=-7/9

c: =>1/2x-1/3-2/3x-1=x

=>-1/6x-4/3=x

=>-7/6x=4/3

=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7

d: =>1-2x-3x+1=7/2

=>-5x=3/2

=>x=-3/10

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

=>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