Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Kết quả là \(\dfrac{101}{299}\). Cô mình chữa rồi đó, và mình lúc đầu cũng không làm được.
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 :
\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+................+\dfrac{1}{299.400}\)
\(101B=\dfrac{101}{1.102}+\dfrac{101}{2.103}+\dfrac{101}{3.104}+...............+\dfrac{101}{299.400}\)
\(101B=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+..................+\dfrac{1}{299}-\dfrac{1}{400}\)
\(101B=\left(1+\dfrac{1}{2}+..............+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}-\dfrac{1}{103}+...............+\dfrac{1}{400}\right)=C\)
\(\Rightarrow B=\dfrac{C}{101}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{C}{101}:\dfrac{C}{299}=\dfrac{101}{299}\)
~ Chúc bn học tốt ~
\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{3012}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\frac{399}{400}\)
\(A=\frac{399}{119600}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\frac{399}{400}\)
\(B=\frac{399}{40400}\)
\(\Rightarrow\frac{A}{B}=\frac{399}{\frac{119600}{\frac{399}{40400}}}=\frac{101}{299}\)
A=11.300+12.301+13.302+...+1101.400�=11.300+12.301+13.302+...+1101.400
A=1299.(11−1300+12−1301+13−13012+...+1101−1400)�=1299.(11−1300+12−1301+13−13012+...+1101−1400)
A=1299.(11−1400)�=1299.(11−1400)
A=1299.399400�=1299.399400
A=399119600�=399119600
B=11.102+12.103+13.104+...+1299.400�=11.102+12.103+13.104+...+1299.400
B=1101.(11−1102+12−1103+....+1299−1400)�=1101.(11−1102+12−1103+....+1299−1400)
B=1101.(11−1400)�=1101.(11−1400)
B=1101.399400�=1101.399400
B=39940400�=39940400
⇒AB=39911960039940400=101299
A=11.300+12.301+13.302+...+1101.400�=11.300+12.301+13.302+...+1101.400
A=1299.(11−1300+12−1301+13−13012+...+1101−1400)�=1299.(11−1300+12−1301+13−13012+...+1101−1400)
A=1299.(11−1400)�=1299.(11−1400)
A=1299.399400�=1299.399400
A=399119600�=399119600
B=11.102+12.103+13.104+...+1299.400�=11.102+12.103+13.104+...+1299.400
B=1101.(11−1102+12−1103+....+1299−1400)�=1101.(11−1102+12−1103+....+1299−1400)
B=1101.(11−1400)�=1101.(11−1400)
B=1101.399400�=1101.399400
B=39940400�=39940400
⇒AB=39911960039940400=101299
A=11.300+12.301+13.302+...+1101.400�=11.300+12.301+13.302+...+1101.400
A=1299.(11−1300+12−1301+13−13012+...+1101−1400)�=1299.(11−1300+12−1301+13−13012+...+1101−1400)
A=1299.(11−1400)�=1299.(11−1400)
A=1299.399400�=1299.399400
A=399119600�=399119600
B=11.102+12.103+13.104+...+1299.400�=11.102+12.103+13.104+...+1299.400
B=1101.(11−1102+12−1103+....+1299−1400)�=1101.(11−1102+12−1103+....+1299−1400)
B=1101.(11−1400)�=1101.(11−1400)
B=1101.399400�=1101.399400
B=39940400�=39940400
⇒AB=39911960039940400=101299
Ta có: \(B=\dfrac{1}{1\cdot102}+\dfrac{1}{2\cdot103}+...+\dfrac{1}{299\cdot400}\)
\(=\dfrac{1}{101}\left(\dfrac{101}{1\cdot102}+\dfrac{101}{2\cdot103}+...+\dfrac{101}{299\cdot400}\right)\)
\(=\dfrac{1}{101}\left(1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+...+\dfrac{1}{299}-\dfrac{1}{400}\right)\)
\(=\dfrac{1}{101}\left(1+\dfrac{1}{2}+...+\dfrac{1}{299}-\dfrac{1}{102}-\dfrac{1}{103}-...-\dfrac{1}{400}\right)\)
\(=\dfrac{1}{101}\left(1+\dfrac{1}{2}+...+\dfrac{1}{101}-\dfrac{1}{300}-\dfrac{1}{301}-...-\dfrac{1}{400}\right)\)
Ta có: \(A=\dfrac{1}{1\cdot300}+\dfrac{1}{2\cdot301}+...+\dfrac{1}{101\cdot400}\)
\(=\dfrac{1}{299}\left(\dfrac{299}{1\cdot300}+\dfrac{299}{2\cdot301}+...+\dfrac{299}{101\cdot400}\right)\)
\(=\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)
\(=\dfrac{1}{299}\left(1+\dfrac{1}{2}+...+\dfrac{1}{101}-\dfrac{1}{300}-\dfrac{1}{301}-...-\dfrac{1}{400}\right)\)
Do đó: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{299}\left(1+\dfrac{1}{2}+...+\dfrac{1}{101}-\dfrac{1}{300}-\dfrac{1}{301}-...-\dfrac{1}{400}\right)}{\dfrac{1}{101}\left(1+\dfrac{1}{2}+...+\dfrac{1}{101}-\dfrac{1}{300}-\dfrac{1}{301}-...-\dfrac{1}{400}\right)}\)
\(=\dfrac{1}{299}\cdot101=\dfrac{101}{299}\)
đề bài sai r nhé bạn