cái lịt

có học thì mới có ăn không làm mà đòi có điểm chỉ có ăn ...

S= 1/199 + 2/198 + ... + 198/2 + 199/1

S= (1/199 + 1) + (2/198 + 1)+ ... + (198/2  + 1) +1

S= 200/200 + 200/199 + 200/198 + ... + 200/2 

S= 200.(1/200 + 1/199 + ... + 1/2)

Suy ra , B=(1/2 + 1/3 + ... +1/200) : 200.(1/2 + 1/3 + ... + 1/200)

B=1 : 200 = 1/200

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ngu vậy chết mm đi

Ta có:

\(A=\frac{1}{1.300}+\frac{1}{2.301}+...+\frac{1}{101.400}\)

\(\Rightarrow A=\frac{1}{299}.\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)

\(\Rightarrow A=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)

\(\Rightarrow A=\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)

Lại có:

\(B=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)

\(\Rightarrow B=\frac{1}{101}.\left(\frac{101}{1.102}+\frac{101}{2.103}+...+\frac{101}{299.400}\right)\)

\(\Rightarrow B=\frac{1}{101}.\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)

\(\Rightarrow B=\frac{1}{101}.\left[\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)\right]\)

\(\Rightarrow B=\frac{1}{101}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}{\frac{1}{101}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}\)

\(\Rightarrow\frac{A}{B}=\frac{1}{299}:\frac{1}{101}\)

\(\Rightarrow\frac{A}{B}=\frac{101}{299}.\)

Vậy \(\frac{A}{B}=\frac{101}{299}.\)

Chúc bạn học tốt!