\(2\dfrac{1}{3}+3\dfrac{1}{2}\)
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a) [-125:(50-75)].(-37)

= {-125: (-25)]. (-37)

= 5. (-37)

= - 185

B)

\(\left(2\dfrac{1}{3}+3\dfrac{1}{2}\right):\left(-4\dfrac{1}{6}+3\dfrac{1}{7}\right)+7\dfrac{1}{2}\\ =\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\\ =\dfrac{35}{6}:\dfrac{-43}{42}+\dfrac{15}{2}\\ =-\dfrac{245}{43}+\dfrac{15}{2}\\ =\dfrac{155}{86}=1\dfrac{69}{86}\)

2 tháng 4 2017

B)\(-\dfrac{55}{58}\)

A)-185

5 tháng 5 2018

dài vậy?Ghi đáp án thôi nhé!hiha

6 tháng 5 2018

1like

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

22 tháng 4 2018

giúp mình đi mà ToT khocroi

a: \(=\dfrac{4\cdot2+4\cdot9}{55}+\dfrac{5}{6}=\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{49}{30}\)

b: \(=\dfrac{3}{2}\cdot\dfrac{3}{5}-\left(\dfrac{3}{7}+\dfrac{3}{20}\right)\cdot\dfrac{10}{3}\)

\(=\dfrac{9}{10}-\dfrac{81}{140}\cdot\dfrac{10}{3}\)

\(=\dfrac{9}{10}-\dfrac{27}{14}=\dfrac{-36}{35}\)

c: \(=15+\dfrac{3}{13}-3-\dfrac{4}{7}-8-\dfrac{3}{13}\)

\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)

d: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=5\)

7 tháng 7 2017

Các câu dễ tự làm nha:

\(D=\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(D=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)\(D=-\dfrac{1}{100}-1\)

10 tháng 12 2022

a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)

b: =(-4)+(-4)+...+(-4)

=-4*25=-100

c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)

=10*53

=530

30 tháng 4 2017

Bài 1:

a) \(\dfrac{2}{5}\cdot x-\dfrac{1}{4}=\dfrac{1}{10}\)

\(\dfrac{2}{5}\cdot x=\dfrac{1}{10}+\dfrac{1}{4}\)

\(\dfrac{2}{5}\cdot x=\dfrac{7}{20}\)

\(x=\dfrac{7}{20}:\dfrac{2}{5}\)

\(x=\dfrac{7}{8}\)

Vậy \(x=\dfrac{7}{8}\).

b) \(\dfrac{3}{5}=\dfrac{24}{x}\)

\(x=\dfrac{5\cdot24}{3}\)

\(x=40\)

Vậy \(x=40\).

c) \(\left(2x-3\right)^2=16\)

\(\left(2x-3\right)^2=4^2\)

\(\circledast\)TH1: \(2x-3=4\\ 2x=4+3\\ 2x=7\\ x=\dfrac{7}{2}\)

\(\circledast\)TH2: \(2x-3=-4\\ 2x=-4+3\\ 2x=-1\\ x=\dfrac{-1}{2}\)

Vậy \(x\in\left\{\dfrac{7}{2};\dfrac{-1}{2}\right\}\).

Bài 2:

a) \(25\%-4\dfrac{2}{5}+0.3:\dfrac{6}{5}\)

\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}:\dfrac{6}{5}\)

\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}\cdot\dfrac{5}{6}\)

\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{1}{4}\)

\(=\dfrac{5}{20}-\dfrac{88}{20}+\dfrac{5}{20}\)

\(=\dfrac{5-88+5}{20}\)

\(=\dfrac{78}{20}=\dfrac{39}{10}\)

b) \(\left(\dfrac{1}{6}-\dfrac{1}{5^2}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=\left(\dfrac{1}{6}-\dfrac{1}{25}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=\left(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=\left(\dfrac{5}{30}-\dfrac{6}{30}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=\left(\dfrac{5-6+1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=0\cdot\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=0\)

Bài 3:

a) \(\dfrac{4}{19}\cdot\dfrac{-3}{7}+\dfrac{-3}{7}\cdot\dfrac{15}{19}\)

\(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)\)

\(=\dfrac{-3}{7}\cdot1\)

\(=\dfrac{-3}{7}\)

b) \(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)

\(=\dfrac{68}{9}-\dfrac{11}{4}-\dfrac{32}{9}\)

\(=\dfrac{68}{9}-\dfrac{32}{9}-\dfrac{11}{4}\)

\(=4-\dfrac{11}{4}\)

\(=\dfrac{16}{4}-\dfrac{11}{4}\)

\(\dfrac{5}{4}\)

Bài 4:

\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)

\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)

\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)

\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)

\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)

\(=2\cdot\dfrac{1}{15}\)

\(=\dfrac{2}{15}\)

Bài 1: 

a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)

b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)

\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)

c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)

\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)