Sửa đề: \(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+...+\frac{1}{299\cdot400}\)
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\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+...+\frac{1}{101\cdot400}\\ A=\frac{1}{299}\left(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+...+\frac{299}{101\cdot400}\right)\\ A=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\\ A=\frac{1}{299}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+\frac{1}{302}+...+\frac{1}{400}\right)\right]\left(1\right)\)
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+...+\frac{1}{299\cdot400}\\ B=\frac{1}{101}\left(\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+...+\frac{101}{299\cdot400}\right)\\ B=\frac{1}{101}\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\\ B=\frac{1}{101}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)\right]\\ B=\frac{1}{101}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}{\frac{1}{101}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}\\ \frac{A}{B}=\frac{\frac{1}{299}}{\frac{1}{101}}=\frac{1}{299}\cdot101=\frac{101}{299}\)
