a, tổng các tử và mẫu mỗi phân sô trên đều bằng 200

b, \(A=\dfrac{1}{199}+\dfrac{2}{198}+\dfrac{3}{197}+...+\dfrac{198}{2}+\dfrac{199}{1}\)

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

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

Ta có:\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)

Lại có:

\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{100}{101}\)

Vậy ...

Những dãy trên đều có 100 số hạng.

Ta có :

\(\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{198}{2}+\dfrac{199}{1}\)

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

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

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

\(=200.A\)

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

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)

\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\)

\(C=\left(\dfrac{1}{200^2}-1\right)\left(\dfrac{1}{199^2-1}\right)...\left(\dfrac{1}{101^2-1}\right)\)

\(C=\dfrac{1-200^2}{200^2}.\dfrac{1-199^2}{199^2}.\dfrac{1-198^2}{198^2}...\dfrac{1-101^2}{101^2}\)

\(C=\dfrac{\left(1-200\right)\left(1+200\right)}{200^2}.\dfrac{\left(1-199\right)\left(1+199\right)}{199^2}...\dfrac{\left(1-100\right)\left(1+100\right)}{100^2}.\dfrac{\left(1-101\right)\left(1+101\right)}{101^2}\) \(C=\dfrac{-199.201}{200.200}.\dfrac{-198.200}{199.199}.\dfrac{-197.199}{198.198}...\dfrac{-99.101}{100.100}.\dfrac{-100.102}{101.101}\)

\(C=\dfrac{199.201}{200.200}.\dfrac{198.200}{199.199}.\dfrac{197.199}{198.198}...\dfrac{99.101}{100.100}.\dfrac{100.102}{101.101}\)

\(\Rightarrow C=\dfrac{200}{2.101}=\dfrac{201}{202}\)

Câu 2 mik chịu r sorry:(

Ta có:  \(\dfrac{1}{101}>\dfrac{1}{200}\)

Tương tự ta có: \(\dfrac{1}{102}>\dfrac{1}{200}\) ;....; \(\dfrac{1}{199}>\dfrac{1}{200}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{200}.100\)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{100}{200}\)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{2}\left(đpcm\right)\)

Bài 1:

a: Sửa đề: 1/3^200

1/2^300=(1/8)^100

1/3^200=(1/9)^100

mà 1/8>1/9

nên 1/2^300>1/3^200

b: 1/5^199>1/5^200=1/25^100

1/3^300=1/27^100

mà 25^100<27^100

nên 1/5^199>1/3^300