=> 3A = 1.2. (3-0) +2.3.(4-1) + ...+18.19.(20-17)

=1.2.3-0.1.2+2.3.4-1.2.3+...+18.19.20-17.18.19

=(1.2.3-1.2.3)+(2.3.4-2.3.4)+...+(17.18.19-17.18.19)+18.19.20-0.1.2

=0+0+0+...+0+18.19.20

=18.19.20

=> A = 6.19.20= 114* 20= 2280

`A= 1.2 + 2.3 +3.4 +...+ 18.19`

`3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +....+ 18.19.(20-17)`

`3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ....+ 18.19.20 - 17.18.19 `

`3A = 18.19.20`

`A = 6.19.20`

`A = 2280`

Đặt `A=1.2+2.3+...+18.19`

`3A=1.2(3-0)+2.3(4-1)+...+18.19(20-17)`

`3A=1.2.3-0.1.2+2.3.4-1.2.3+...+18.19.20-17.18.19`

`3A=18.19.20=6840`

\(\Rightarrow\) `A=6840/3`

\(\Rightarrow\) `A=2280`

`@Nae`

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

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

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

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

\(\Rightarrow A=\frac{19}{10}\)

2.(1/1.2+1/2.3+.....+1/18.19+1/19.20)

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

2.(1/1-1/20)= 2.19/20=19/10

Mỏi tay nhắm . không viết đầu bài đâu >.<

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

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

= \(2\left(1-\frac{1}{20}\right)\)

= \(2.\frac{19}{20}\)

= \(\frac{19}{10}\)

Mị không chắc đâu :3

\(A=1.2+2.3+...+18.19\)

\(3A=1.2\left(3-0\right)+2.3\left(4-1\right)+...+18.19\left(20-17\right)\)

\(3A=1.2.3-0.1.2+2.3.4-1.2.3+...+18.19.20-17.18.19\)

\(3A=18.19.20=6840\)

\(\Rightarrow A=\dfrac{6840}{3}=2280\)

=2919

dạng tổng quát của mỗi phân số là 1/n(n+1) = 1/n -1/n+1

áp dụng vào làm với các phân số trong biểu thức cuối cùng còn 1-1/10=19/20

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

=  \(\frac{19}{10}\)

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

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

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

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

Lời giải:
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+....+\frac{19-18}{18.19}+\frac{20-19}{19.20}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}$

$=1-\frac{1}{20}=\frac{19}{20}$

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

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

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

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

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

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

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

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

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

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

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

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