:2 = 1/6 + 1/12 + 1/20 +...+ 1/380
= 1/(2x3) + 1/(3x4) + 1/(4x5) + ... + 1/(19x20)
= 1/2 - 1/3+1/3 - 1/4 +....+ 1/19- 1/20
= 1/2 - 1/20 = 9/20
Suy ra A = 9/20 x 2 = 9/10

A:2 = 1/6 + 1/12 + 1/20 +...+ 1/380 = 1/(2x3) + 1/(3x4) + 1/(4x5) + ... + 1/(19x20) = 1/2 - 1/3+1/3 - 1/4 +....+ 1/19- 1/20 = 1/2 - 1/20 = 9/20 Suy ra A = 9/20 x 2 = 9/10

S=2(1/2+1/6+...+1/380)

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

=2*19/20=19/10

Ta viết lại dãy phân số như sau: 
1/(0+1) + 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) + ......+ 1/(1+2+3+ ....+19) 
= 1/2x(0+1):2 + 2/2x(1+3) + 2/3x(1+3) + 2/4x(1+4) + ......+ 2/19x(1+19) 
= 2/1.2 + 2/2.3 + 2/3.4 + .....+2/19.20 (thay dấu x bằng dấu chấm cho đỡ rối) 
= 2.( 1-1/2 + 1/2 - 1/3 + 1/3 -1/4 + ....+ 1/19 -1/20) 
= 2.( 1- 1/20) 
= 2. 19/20 
= 19/10 
= 1 + 9/10 
có thể để kết quả là 19/10 hay đổi ra hỗn số tùy ý.

sửa đề nè : 1+ 1\3+1\6+1\10+...+1\190

nè sửa đi chứ thêm số 1 vào

Ta có \(\left|x+1\right|\ge0;\left|x+\frac{1}{3}\right|\ge0;...;\)\(\left|x+\frac{1}{190}\right|\ge0\)    \(\forall x\)

=> \(\left|x+1\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{190}\right|\ge0\)   \(\forall x\)

=> \(20x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\)  =>  \(x+1>0,x+\frac{1}{3}>0,x+\frac{1}{6}>0,...,x+\frac{1}{190}>0\)

                        => \(\left|x+1\right|=x+1,\left|x+\frac{1}{3}\right|=x+\frac{1}{3},\left|x+\frac{1}{6}\right|=x+\frac{1}{6},...,\left|x+\frac{1}{190}\right|=x+\frac{1}{190}\)

=> \(x+1+x+\frac{1}{3}+x+\frac{1}{6}+...+x+\frac{1}{190}=20x\)

=> \(19x+\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)=20x\)

=> \(x=\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)\)

Gọi \(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{190}\)

=> \(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)

=> \(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

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

=> \(\frac{1}{2}A=1-\frac{1}{20}\)

=> \(A=\frac{19}{10}\)

Thay vào ta có 

=> \(x=-\frac{19}{10}\)

mk nhầm nha bạn \(x=\frac{19}{10}\)

Đặt \(M=68+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{190}\)

Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{190}\)

\(\Rightarrow A\times\frac{1}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{380}\)

\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{19\times20}\)

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

\(=1-\frac{1}{20}=\frac{19}{20}\)

\(\Rightarrow M=68+\frac{19}{20}=\frac{1379}{20}\)

Vậy \(68+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{190}=\frac{1379}{20}\)

Cho tớ sửa lại dòng thứ 2 từ dưới lên :

\(\Rightarrow A=\frac{19}{20}:\frac{1}{2}=\frac{19}{10}\)

\(\Rightarrow M=68+\frac{19}{10}=\frac{699}{10}\)

Vậy ...

Bài hỏi j?

=2*(1/6+1/12+1/20+...+1/380)

=2*(1/2*3+1/3*4+1/4*5+...+1/19*20)

=2*(1/2-1/3+1/3-1/4+1/4-1/5+...+1/19-1/20)

=2*(1/2-1/20)

=2*(10/20-1/20)

=2*9/20

=18/20

=9/10

A/2=(1/3+1/6+1/10+1/15+...+1/171+1/190)/2

=1/6+1/12+1/20+1/30+...+1/342+1/380

=1/2.3+1/3.4+1/4.5+1/5.6+....+1/18.19+1/19.20

=1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 +....+ 1/18 - 1/19 + 1/19 - 1/20

=1/2 - 1/20 = 10/20 - 1/20 = 9/20

=> A = 9/20 .2 = 18/20=9/10