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19 tháng 3 2016

câu hỏi?

17 tháng 6 2015

Ta có:

1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9+1/9*10

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

=1-1/10

=9/10

17 tháng 6 2015

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)

25 tháng 6 2018

\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)

\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)

\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)

\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)

\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)

\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)

\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)

\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)

\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)

18 tháng 2 2021

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7 tháng 8 2018

=1/1x2 + 1/2x3 + 1/3x4 +1/4x5+ 1/5x6 + 1/6x7 + 1/7x8 + 1/8x9 + 1/9x10

=1/1 -1/2 +1/2 -1/3 +1/3 -1/4+1/4-1/5+ 1/5 -1/6 + 1/6-1/7 + 1/7-1/8 + 1/8- 1/9 + 1/9- 1/10

=1/1 -1/10

=9/10

4 tháng 1 2022

bằng 131/840=0,155952381

17 tháng 6 2017

a)y=5/6

b)y=11/4

17 tháng 6 2017

\(a\left(\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2}-\frac{1}{10}\right).y=\frac{1}{3}\)

\(\frac{2}{5}.y=\frac{1}{3}\)

      \(y=\frac{1}{3}:\frac{2}{5}\)

     \(y=\frac{5}{6}\)

\(b,\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

     \(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

      \(\frac{10}{11}.y=\frac{2}{3}\)

              \(y=\frac{2}{3}:\frac{10}{11}\)

               \(y=\frac{22}{30}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{11}{33}-\dfrac{35}{40}\)

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

`=`\(\dfrac{12}{24}-\dfrac{20}{24}+\dfrac{8}{24}-\dfrac{21}{24}\)

`= -21/24 = -7/8`

`b)`

\(\dfrac{2}{3}\cdot1\dfrac{3}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{2}{3}\cdot\dfrac{7}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{7}{6}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{5}{18}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(-\dfrac{1}{18}-\dfrac{1}{5}=-\dfrac{23}{90}\)

`c)`

\(\dfrac{1}{2}\cdot2-2\dfrac{5}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(1-\dfrac{19}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(-\dfrac{12}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(-\dfrac{3}{14}-\dfrac{10}{15}=-\dfrac{37}{42}\)

`d) `

\(\dfrac{1}{6}\cdot\dfrac{1}{11}+\dfrac{4}{11}\cdot\left(-\dfrac{1}{6}\right)+\dfrac{8}{11}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{6}{11}\)

`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1}{11}-\dfrac{4}{11}+\dfrac{8}{11}+\dfrac{6}{11}\right)\)

`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1-4+8+6}{11}\right)\)

`=`\(\dfrac{1}{6}\cdot1=\dfrac{1}{6}\)

`e)`

\(-17\cdot\left(-23\right)+\left(-53\right)\cdot17+17\cdot14+17\cdot\left(-24\right)\)

`= 17*(23-53+14-24)`

`= 17*(-40)`

`= -680`

`f)`

\(-19\cdot218+\left(-82\right)\cdot19-533\cdot19+\left(-19\right)\cdot167\)

`= 19*(-218-82-533-167)`

`= 19*(-1000)`

`= -19000`

`g)`

\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{11}{44}+\dfrac{9}{16}\)

`=`\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{9}{16}\)

`=`\(\dfrac{31}{40}-\dfrac{1}{4}+\dfrac{9}{16}\)

`=`\(\dfrac{21}{40}+\dfrac{9}{16}=\dfrac{87}{80}\)

`h)`

\(\dfrac{4}{10}-1\dfrac{5}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{4}{10}-\dfrac{11}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{4}{10}-\dfrac{11}{3}+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(-\dfrac{49}{15}+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(-\dfrac{287}{120}-\dfrac{1}{9}=-\dfrac{901}{360}\)

`i )`

\(3\cdot\dfrac{1}{5}-\dfrac{2}{8}-\dfrac{12}{36}+\dfrac{15}{9}\)

`=`\(\dfrac{3}{5}-\dfrac{1}{4}-\dfrac{1}{3}+\dfrac{15}{9}\)

`=`\(\dfrac{7}{20}-\dfrac{1}{3}+\dfrac{15}{9}\)

`=`\(\dfrac{1}{60}+\dfrac{15}{9}=-\dfrac{33}{20}\)

`k)`

\(\dfrac{6}{8}\cdot3\dfrac{1}{2}+4\dfrac{2}{3}-\dfrac{11}{55}+\dfrac{17}{51}\)

`=`\(\dfrac{3}{4}\cdot\dfrac{7}{2}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{21}{8}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{175}{24}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{851}{120}+\dfrac{17}{51}=\dfrac{297}{40}\)

`l )`

\(\dfrac{1}{3}\cdot3\dfrac{1}{2}-4\dfrac{2}{5}-\dfrac{26}{78}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\cdot\dfrac{7}{2}-\dfrac{22}{5}-\dfrac{1}{3}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\left(\dfrac{7}{2}-1\right)-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\cdot\dfrac{5}{2}-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{5}{6}-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(-\dfrac{107}{30}+\dfrac{17}{51}=-\dfrac{97}{30}\)

P/s: Bạn tách bài ra hỏi nhé! Và ghi đề rõ ràng chứ đừng ghi ntnay, nhiều bạn nhìn vào rất khó nhìn!

`# \text {KaizulvG}`

GH
26 tháng 6 2023

27 tháng 6 2018

1/ \(x+\dfrac{1}{2}=\dfrac{-5}{3}\)

\(x=\dfrac{-5}{3}-\dfrac{1}{2}\)

\(x=\dfrac{-10}{6}-\dfrac{3}{6}\)

Vậy \(x=\dfrac{-13}{6}\)

2/\(\dfrac{1}{3}-x=\dfrac{3}{5}\)

\(-x=\dfrac{3}{5}-\dfrac{1}{3}\)

\(-x=\dfrac{9}{15}-\dfrac{5}{15}\)

\(-x=\dfrac{4}{15}\)

Vậy \(x=\dfrac{-4}{15}\)

3/ \(3-4+x=\dfrac{7}{2}\)

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

\(-4+x=\dfrac{7}{2}-\dfrac{6}{2}\)

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

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

\(x=\dfrac{1}{2}+\dfrac{8}{2}\)

Vậy \(x=\dfrac{9}{2}\)

4/ \(x-\dfrac{4}{3}=\dfrac{-7}{9}\)

\(x=\dfrac{-7}{9}+\dfrac{4}{3}\)

\(x=\dfrac{-7}{9}+\dfrac{12}{9}\)

Vậy \(x=\dfrac{5}{9}\)

5/ \(x-\dfrac{-7}{2}=\dfrac{5}{6}\)

\(x=\dfrac{5}{6}-\dfrac{7}{2}\)

\(x=\dfrac{5}{6}-\dfrac{21}{6}\)

Vậy \(x=\dfrac{-16}{6}=\dfrac{-8}{3}\)

6/ \(x-\dfrac{1}{5}=\dfrac{9}{10}\)

\(x=\dfrac{9}{10}+\dfrac{1}{5}\)

\(x=\dfrac{9}{10}+\dfrac{2}{10}\)

Vậy \(x=\dfrac{11}{10}\)

7/ \(x+\dfrac{5}{12}=\dfrac{3}{8}\)

\(x=\dfrac{3}{8}-\dfrac{5}{12}\)

\(x=\dfrac{9}{24}-\dfrac{10}{24}\)

Vậy \(x=\dfrac{-1}{24}\)

8/ \(x+\dfrac{5}{4}=\dfrac{7}{6}\)

\(x=\dfrac{7}{6}-\dfrac{5}{4}\)

\(x=\dfrac{14}{12}-\dfrac{15}{12}\)

Vậy \(x=\dfrac{-1}{12}\)

9/ \(x-\dfrac{2}{7}=\dfrac{1}{35}\)

\(x=\dfrac{1}{35}+\dfrac{2}{7}\)

\(x=\dfrac{1}{35}+\dfrac{10}{35}\)

Vậy \(x=\dfrac{11}{35}\)

10 /\(x-\dfrac{1}{5}=\dfrac{-7}{10}\)

\(x=\dfrac{-7}{10}+\dfrac{1}{5}\)

\(x=\dfrac{-7}{10}+\dfrac{2}{10}\)

Vậy \(x=\dfrac{-5}{10}=\dfrac{-1}{2}\)