A = 1 - 2 + 3 - 4 + ... + 59 - 60

A = (1 - 2) + (3 - 4) + ... + (59 - 60)

A = -1 + (-1) + ... + (-1)            có 30 số -1  

A = -1.30

A = -30

\(B=\frac{-3}{20}-\frac{3}{200}-\frac{3}{2000}-\frac{3}{20000}\)

\(B=\frac{-3000}{20000}-\frac{300}{20000}-\frac{30}{20000}-\frac{3}{20000}\)

\(B=\frac{-3333}{20000}\)

Mk chỉ làm đc phần a thui nha bạn !

\(A=1-2+3-4+...+59-60\)

\(A=\left(1-2\right)+\left(3-4\right)+...+\left(59-60\right)\)

\(A=-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)

                 Có tổng cộng 30 số \(\left(-1\right)\)

\(A=30.\left(-1\right)\)

\(A=-30\)

Dii vào link nào nha : https://loga.vn/hoi-dap/tinh-tong-s-3-1-2-3-2-3-3-3-4-3-4-5-3-2015-20161-tinh-tong-s-dfrac-3-1-2-dfrac-3-2-3-dfrac-3-3-4-17371 

k mk

Đặt: \(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{199}{1}\)là B

Cộng 1 vào mỗi phần số trừ phân số cuối cùng ta sẽ được:

B= \(\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)

=> B= \(\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+1\)

=> B= \(\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+\frac{200}{200}\)

=> B= \(200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)

Đặt \(A=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\) => B= \(200\) X A

=> \(\frac{A}{B}\)\(=\frac{1}{200}\)

=> \(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)

=>\(x-20\) =\(\frac{1}{2000}:\frac{1}{200}\)

=> \(x-20=\).......................... Bạn tự làm tiếp nhé, chúc bạn học tốt !!!^^\(\)

 \(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)

\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+....+\left(\frac{198}{2}+1\right)+1}=\frac{1}{2000}\)

\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}}=\frac{1}{2000}\)

\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{2000}\)

\(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)

\(\left(x-20\right)=\frac{1}{2000}:\frac{1}{200}=\frac{1}{2000}.200=\frac{1}{10}\)

\(\Rightarrow x=\frac{1}{10}+20=\frac{201}{10}\)

= 1/10

Ta có:\(\frac{\frac{20}{200}}{2000}=\frac{\frac{1}{10}}{2000}=\frac{1}{10}:\frac{1}{2000}=\frac{1}{10}.2000=200\)

Vậy bạn HS A trả lời đúng

20/200/2000=1/10/2000=1/10:1/2000=1/10.2000=2000

a)      Ta có: \(\frac{2}{{ - 5}} = \frac{{ - 16}}{{40}}\) và \(\frac{{ - 3}}{8} = \frac{{ - 15}}{{40}}\)

Do \(\frac{{ - 16}}{{40}} < \frac{{ - 15}}{{40}}\,\, \Rightarrow \,\frac{2}{{ - 5}} < \frac{{ - 3}}{8}\).

b)      Ta có: \( - 0,85 = \frac{{ - 85}}{{100}} = \frac{{ - 17}}{{20}}\). Vậy \( - 0,85\)=\(\frac{{ - 17}}{{20}}\).

c)      Ta có: \(\frac{{37}}{{ - 25}} = \frac{{ - 296}}{{200}}\)  

Do  \(\frac{{ - 137}}{{200}} > \frac{{ - 296}}{{200}}\) nên \(\frac{{ - 137}}{{200}}\) > \(\frac{{37}}{{ - 25}}\) .

d)      Ta có: \( - 1\frac{3}{{10}}=\frac{-13}{10}\) ;

\(-\left( {\frac{{ - 13}}{{ - 10}}} \right) = \frac{{-13}}{{10}}\).

Vậy \(- 1\frac{3}{{10}} =-(\frac{{-13}}{{-10}})\,\).

\(\left[\frac{2000}{2000.2006}+\frac{2000}{2006.2012}+...+\frac{2000}{2492.2498}\right]\times\left[\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\right]\)

\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2006}+...+\frac{1}{2492}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{8.11}+\frac{9}{11.14}+...+\frac{9}{197.200}\right]\)

\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+..+\frac{1}{197}-\frac{1}{200}\right)\right]\)

\(=\left[\frac{2000}{6}\cdot\frac{498}{4996000}\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{200}\right)\right]\)

\(=\frac{83}{2498}\times\left[\frac{9}{3}\cdot\frac{3}{25}\right]\)

\(=\frac{83}{2498}\times\frac{9}{25}=\frac{747}{62450}\)