Cho a,b,c \(\in\)N* và a<b<1.Ta có:\(\frac{a}{b}<\frac{a+c}{b+c}\)

\(\Rightarrow\)a(b+c)<b(a+c)

\(\Rightarrow\)ab+ac<ba+bc

\(\Rightarrow\)ac<bc

Tiếp nè:

\(\Rightarrow\)a<b đúng

Mặt khác:\(\frac{1}{2}<\frac{1+1}{2+1}=\frac{2}{3}\)

              \(\frac{3}{4}<\frac{3+1}{4+1}=\frac{4}{5}\)

               \(\frac{199}{200}<\frac{199+1}{200+1}=\frac{200}{201}\)

\(\Rightarrow A<\frac{2}{3}.\frac{4}{5}...........\frac{200}{201}\)

\(\Rightarrow A^2<\frac{1}{2}.\frac{2}{3}.\frac{3}{4}............\frac{199}{200}.\frac{200}{201}\)

\(\Rightarrow A^2<\frac{1}{101}<\frac{1}{100}\)

\(\Rightarrow A<\frac{1}{10}\)

b,Chưa làm được,sorry

A=1/1.2+1/2.3+1/3.4+... tu do tu lam nhe

- Đây http://olm.vn/hoi-dap/question/89634.html

196/197 + 197/198 > 198/199 + 199/197

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

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

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

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

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

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

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

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

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

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

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

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

\(=200\left(\frac{1}{100}+\frac{1}{199}+\frac{1}{198}+....+\frac{1}{2}\right)\)

= 200.A

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

\(A=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{189}{2}+\frac{199}{1}\)

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

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

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

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

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

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