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

\(=1+\frac{1}{199}+1+\frac{2}{198}+...+\frac{199}{1}+1-199\)

\(=200+\frac{200}{2}+...+\frac{200}{199}-199\)

\(=1+\frac{200}{2}+...+\frac{200}{199}\)

\(=200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{200}\)

\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)

\(=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+...+\left(199-1-1-1-...1\right)\)(198 chữ số 1)

\(=\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+1=200.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{197}+\frac{1}{198}+\frac{1}{199}+\frac{1}{200}\right)=200.A\)

\(\Rightarrow\frac{A}{B}=\frac{A}{200.A}=\frac{1}{200}\)

i don't now

mong thông cảm !

...........................

B = \(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)

\(=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+199\)

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

(từ 1 đến 198 có 198 số hạng nên còn 1 số 1)

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

\(=200\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{\text{4}}+...+\frac{1}{200}\right)=200A\)

=> B = 200A => \(\frac{A}{B}=\frac{1}{200}\)

Vậy \(\frac{A}{B}=\frac{1}{200}\)

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)

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

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+\frac{200}{200}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{200\cdot\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right)}\)

\(=\frac{1}{200}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\left[\frac{1}{199}+1\right]+\left[\frac{2}{198}+1\right]+\left[\frac{3}{197}+1\right]+...+\left[\frac{198}{2}+1\right]}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{200\left[\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right]}=\frac{1}{200}\)

Ta có : 

\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)

\(B=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+...+\left(\frac{198}{2}+1\right)+\left(\frac{199}{1}-1-1-1-...-1\right)\)

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

\(B=200\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(\Rightarrow\)\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)}=\frac{1}{200}\)

Vậy \(\frac{A}{B}=\frac{1}{200}\)

Chúc bạn học tốt ~ 

\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)   (1)

\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)     (2)

\(\left(1\right)\left(2\right)\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{198}+1\right)+...+\left(\frac{198}{2}+1\right)+1}\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+\frac{200}{200}}\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{200}\right)}\)

\(\Rightarrow\frac{A}{B}=\frac{1}{200}\)

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 !!!^^\(\)

Bn kham khảo câu hỏi tương t5uwj nha

chúc bn học tốt 

nhớ k mình nha

cám ơn bn 

và bn cũng có thể kham khảo học 24 nha 

tôi nghĩ bài này biến đổi đc  thui chứ ko tính đc.

\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....+\frac{198}{2}+\frac{199}{1}\)

\(=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+....+\left(\frac{199}{1}+1\right)-199\)

\(=\frac{200}{199}+\frac{200}{198}+....+\frac{200}{1}+\frac{200}{200}-200\)

\(=200.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)\)