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9 tháng 7 2017

3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)

\(\Leftrightarrow2x-6-3+6x=4+4-4x\)

\(\Leftrightarrow8x-9=8-4x\)

\(\Leftrightarrow8x+4x=8+9\)

\(\Leftrightarrow12x=17\)

\(\Leftrightarrow x=\dfrac{17}{12}\)

Vậy \(x=\dfrac{17}{12}\)

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

\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)

\(\Leftrightarrow6x-12-4-4x=12-9x-12\)

\(\Leftrightarrow6x-4-4x=12-9x\)

\(\Leftrightarrow2x-4=12-9x\)

\(\Leftrightarrow2x+9x=12+4\)

\(\Leftrightarrow11x=16\)

\(\Leftrightarrow x=\dfrac{16}{11}\)

Vậy \(x=\dfrac{16}{11}\)

6 tháng 4 2018

Ta có: \(\frac{-3}{1.2.3}+\frac{-3}{2.3.4}+\frac{-3}{3.4.5}+...+\frac{-3}{18.19.20}\)

          \(=\frac{-3}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{18.19.20}\right)\)

          \(=\frac{-3}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

            \(=\frac{-3}{2}\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{-3}{2}.\frac{189}{380}=\frac{-567}{760}\)

              

22 tháng 6 2017

a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)

\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)

\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{13}{2}\cdot\frac{8}{33}\)

\(=\frac{52}{33}\)

22 tháng 6 2017

a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99

A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)

A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)

A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)

A= 13/2 ( 1/3 - 1/11) 

A= 13/2 . 8/33

A= 52/33  

20 tháng 6 2016

http://olm.vn/tin-tuc/Bai-toan-106.html

tham khảo nhé, vội quá ko trình bày cho bạn được

14 tháng 4 2017

99/100 k nha

12 tháng 3 2018

999/100

12 tháng 7 2017

B= 1/ 1.2.3 + 1/ 2.3 4 + 1/ 3.4.5 + .... + 1/ 18.19.20

Ta có:

1/ 1.2 - 1/ 2.3 = 2/ 1.2.3

1/ 2.3 - 1/3.4 = 2/ 2.3.4

Từ đó Ta có: B = 1/2 . ( 2/ 1.2.3 + 2/ 2,3.4 + ... + 2/ 18. 19. 20 )

= 1/2 .( 1/ 1.2 – 1/ 2.3 + 1/ 2.3 - .....- 1/19.20)

= 1/2. ( 1/ 1.2 – 1/ 19.20 ) = 1/ 2 . 189/380 = 189/760

13 tháng 7 2017

\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+....+\frac{1}{18\cdot19\cdot20}\)

\(B=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{20-18}{18\cdot19\cdot20}\)

\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{18\cdot19\cdot20}\)

\(2B=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\)

\(2B=\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\)

\(\Rightarrow B=\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\div2=\frac{189}{380}\div2=\frac{189}{760}\)

\(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2021\cdot2022\cdot2023}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{2021\cdot2022\cdot2023}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2021\cdot2022}-\dfrac{1}{2022\cdot2023}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4090506}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2045252}{4090506}=\dfrac{1022626}{4090506}=\dfrac{511313}{2045253}\)

2 tháng 8 2023

`1/(1.2.3) + 1/(2.3.4) + ... + 1/(2021 . 2022 .2023)`

`=> 2/(1.2.3) + 2/(2.3.4) + ... + 2/(2021 . 2022. 2023)`

`= 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + ... + 1/(2021.2022) - 1/(2022 . 2023)`

`= 1/2 - 1/4090506`

`=4090506/8181012 - 2/8181012`

`= 4090504/8181012`

27 tháng 12 2018

Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)

=>2A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+...+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\)\(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\)

=>A=\(\dfrac{n^2+3n}{4n^2+12n+8}\)

23 tháng 8 2023

\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)

24 tháng 8 2023

Thanks