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Ta có :
\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+..................+\dfrac{1}{101.400}\)
\(299A=\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+..................+\dfrac{299}{101.400}\)
\(299A=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+.................+\dfrac{1}{101}-\dfrac{1}{400}\)
\(299A=\left(1+\dfrac{1}{2}+.................+\dfrac{1}{101}\right)-\left(\dfrac{1}{300}+\dfrac{1}{301}+.............+\dfrac{1}{400}\right)=C\)
\(\Rightarrow A=\dfrac{C}{299}\)
Lại có :
\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+................+\dfrac{1}{299.400}\)
\(101B=\dfrac{101}{1.102}+\dfrac{101}{2.103}+...............+\dfrac{101}{299.400}\)
\(101B=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+...............+\dfrac{1}{299}-\dfrac{1}{400}\)
\(101B=\left(1+\dfrac{1}{2}+...............+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+...............+\dfrac{1}{400}\right)=C\)\(\Rightarrow B=\dfrac{C}{101}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{C}{101}:\dfrac{C}{299}=\dfrac{299}{101}\)
~ Chúc bn học tốt ~
\(A=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...........+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left(1+\frac{1}{2}+........+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-....-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+........+\frac{1}{299}-\frac{1}{400}\right)\)
\(=\frac{1}{101}.\left(1+\frac{1}{2}+.......+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-............-\frac{1}{400}\right)\)
\(=\frac{1}{101}\left(1+\frac{1}{2}+......+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{299}-\frac{1}{102}-.....-\frac{1}{300}-....-\frac{1}{400}\right)\)
\(=\frac{1}{101}\left(1+\frac{1}{2}+........+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}}{\frac{1}{101}}=\frac{101}{299}\)
299A=(1+1/2+1/3+...+1/101)-(1/300+1/301+...+1/400)=C
101B=(1+1/2+1/3+...+1/299)-(1/102+1/103+..+1/400)=D=C
=>A/B=C/299.101/C=101/299
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Ta có: \(A=\frac{23}{1\cdot300}+\frac{23}{2\cdot301}+\frac{23}{3\cdot302}+\cdots+\frac{23}{101\cdot400}\)
\(=23\left(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\cdots+\frac{1}{101\cdot400}\right)\)
\(=\frac{23}{299}\left(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+\cdots+\frac{299}{101\cdot400}\right)\)
\(=\frac{23}{299}\left(1-\frac{1}{300}+\frac12-\frac{1}{301}+\cdots+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{13}\left(1+\frac12+\cdots+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\cdots-\frac{1}{400}\right)\)
Ta có: \(B=\frac{1313}{1\cdot102}+\frac{1313}{2\cdot103}+\cdots+\frac{1313}{299\cdot400}\)
\(=13\left(\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+\cdots+\frac{101}{299\cdot400}\right)\)
\(=13\left(1-\frac{1}{102}+\frac12-\frac{1}{103}+\cdots+\frac{1}{299}-\frac{1}{400}\right)\)
\(=13\left(1+\frac12+\cdots+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\cdots-\frac{1}{400}\right)\)
Do đó; \(\frac{A}{B}=\frac{1}{13}:13=\frac{1}{169}\)
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