\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{19.20}-\frac{x}{40}=\frac{3}{-10}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)

\(\Rightarrow1-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)

\(\Rightarrow\frac{40}{40}-\frac{2}{40}-\frac{x}{40}=\frac{-12}{40}\)

\(\Rightarrow\frac{38}{40}-\frac{x}{40}=\frac{-12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{38}{40}-\frac{-12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{38}{40}+\frac{12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{50}{40}\)

\(\Rightarrow x=50\)

Vậy x = 50

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{19\cdot20}-\frac{x}{40}=\frac{-3}{10}\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{19}-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)

\(1-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)

\(\frac{x}{40}=1-\frac{1}{20}-\frac{3}{-10}=1\frac{1}{4}=\frac{5}{4}\)

\(\frac{x}{40}=\frac{5}{4}\Rightarrow x=\frac{40\cdot5}{4}=50\)

Đặt  \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)

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

\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(A=2\left(1-\frac{1}{20}\right)\)

\(A=2.\frac{19}{20}=\frac{19}{10}\)

Vậy ...

=2.(\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+......+\(\frac{1}{19.20}\))

=2.( 1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+..........+\(\frac{1}{19}\)-\(\frac{1}{20}\))

=2.(1-\(\frac{1}{20}\))

=2.\(\frac{19}{20}\)

=  \(\frac{19}{10}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2.\left(1-\frac{1}{20}\right)\)

\(=2\cdot\frac{19}{20}=\frac{19}{10}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{19.20}\)

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

\(=2.\left(1-\frac{1}{20}\right)\)

\(=2.\frac{19}{20}=\frac{19}{10}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{1}-\frac{1}{20}\)

\(=\frac{19}{20}\)

Đây là toán lớp 6 mà bạn. Lớp 5 ở đâu ra có dạng bài toán này

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(A=1-\frac{1}{20}\)

\(A=\frac{19}{20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}\)

\(=\frac{19}{20}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(A=1-\frac{1}{20}\)

\(A=\frac{19}{20}\)

Ta có:A: 1/1.2 +1/2.3 +1/3.4+...+1/18.19+1/19.20
     => A= 1-1/2 +1/2-1/3+1/3-1/4+...+1/18-1/19+1/19-1/20

     =>A= 1-1/20=19/20

a) = 1-1/2+1/2-1/3+1/3-1/4

    = 1-1/4=3/4

b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018

   =1-1/2018=2017/2018

c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015

   = 1/2-1/2015=2015/4030-2/4030=2013/4030

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)

b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)

\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)

\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}.\frac{2013}{4030}\)

\(=\frac{6039}{8060}\)

]\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a) = 2(1-1/2+1/2-1/3+...+1/19-1/20)

    = 2(1-1/20)

    = 2.19/20

    = 19/10

b) = 7(1/2-1/3+1/3-1/4+...+1/6-1/7)

   = 7(1/2 - 1/7)

   = 7.5/14

   = 5/2

c) = 1/2-1/5+1/5-1/8+...+1/14-1/17

   = 1/2 - 1/17

   = 15/34

Chúc bạn học tốt nhé

a)2/1.2+2/2.3+....+2/19.20

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

=2(1-1/2+1/2-1/3+.....-1/20)

=2(1-1/20)

2(19/20)=38/20=19/10

b)7/2.3+7/3.4+7/4.5+7/5.6+7/6.7

7(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7)

7(1/2-1/3+1/3-1/4+.....-1/7)

7(1/2-1/7)

7(7/14-2/14)=7.5/14=35/14=5/2

c)3/2.5+3/5.8+3/8.11+3/11.14+3/14.17

1/2-1/5+1/5-1/8+......+1/14-1/17

1/2-1/17=17/34-2/34=15/34

Đây là toán lp 6 chứ bn

A = 2/1.2 + 2/2.3 + ...+2/19.20

Áp dụng công thức : n/n.(n+1)=(1/n)-(1/n+1)

Ta có : A= 1/1.2 + 1/2.3+...+1/19.20

           A= 1/1 - 1/2 + 1/2 - 1/3 + ...+ 1/19 - 1/20

           A= 1/1 - 1/20 

           A= 19/20

k cho mk nha bn

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)

=1-1/20

=19/20

\(S=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+....+\frac{3}{2015.2016}\)

\(\Rightarrow\frac{1}{3}.S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2015.2016}\)

\(\Rightarrow\frac{1}{3}.S=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+......+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)

\(\Rightarrow\frac{1}{3}.S=\frac{1}{1}-\frac{1}{2016}\)

\(\Rightarrow\frac{1}{3}.S=\frac{2015}{2016}\)

\(\Rightarrow S=\frac{2015}{672}\)

Vậy: \(\Rightarrow S=\frac{2015}{672}\)

Bạn giải giúp mk câu mk đăng tầm 5 phút nha!

đơn giản

S=1/1-1/2+1/2-1/3+1/3-1/4+1\5-..............................-1/2015+1/2016

S=1/1-1/2016

S=2015/2016