tính A:B A=1/1300+1/2301+1/3302+...+1/101400 B=1/1.102+1/2.103+...+1/299.400
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TL
2
MA
11 tháng 9 2020
Ta có: a=1/1.300+1/2.301+...+1/101.400
⇒ a= 1/299.(299/1.300+299/2.301+...+299/101.400)
⇒ a= 1/299. ( 1+1/300+1/2-1/301+....+1/101-1/400)
⇒ a= 1/299.|(1+1/2+....+1/101)-(1/300+1/301+....+1/400)|
Ta có: b=1/1.102+1/2.103+..+1/299.400
⇒ b= 1/101.(101/1.102+101/2.103+..+101/299.400)
⇒ 1/101.|(1-1/102+1/2-1/102+......+1/299-1/400)|
⇒ b= 1/101 .|(1+1/2+....+1/299) - (1/102+1/103+....+1/400)|
⇒ b= |(1+1/2+....+1/299)- (1/300+1/301+....+1/400)|
⇒a=1/299.|(1+1/2+....+1/101)-(1/300+1/301+....+1/400)|
phần
b=1/101.|(1+1/2+....+1/101)-(1/300+1/301+....+1/400)|
⇒a/b=1/299:1/101
⇒a/b=101/299.
11 tháng 9 2020
Ta chú ý đẳng thức \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)(Chứng minh rất dễ, bạn quy đồng lên là được nha)
\(A=\frac{1}{1.300}+\frac{1}{2.301}+...+\frac{1}{101.400}\)
\(\Rightarrow299A=\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101+400}\)
\(=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\)
Đặt \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}=X,\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}=Y\)
\(\Rightarrow A=\frac{X-Y}{299}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(\Rightarrow101B=\frac{101}{1.102}+\frac{101}{2.103}+\frac{101}{3.104}+...+\frac{101}{299.400}\)
\(=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+\frac{1}{3}-\frac{1}{104}+...+\frac{1}{102}-\frac{1}{203}+\frac{1}{103}-\frac{1}{204}+...\)
\(\frac{1}{198}-\frac{1}{299}+\frac{1}{199}-\frac{1}{300}+\frac{1}{200}-\frac{1}{301}+...+\frac{1}{299}-\frac{1}{400}\)
\(=\left(1+...+\frac{1}{101}\right)-\left(\frac{1}{300}+...+\frac{1}{400}\right)+\left(\frac{1}{102}-\frac{1}{102}\right)+\left(\frac{1}{103}-\frac{1}{103}\right)+...+\left(\frac{1}{299}-\frac{1}{299}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)=X-Y\)
\(\Rightarrow B=\frac{X-Y}{101}>\frac{X-Y}{299}=A\)
Vậy \(B>A\)
30 tháng 11 2023
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
DP
25 tháng 6 2017
\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{3012}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\frac{399}{400}\)
\(A=\frac{399}{119600}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\frac{399}{400}\)
\(B=\frac{399}{40400}\)
\(\Rightarrow\frac{A}{B}=\frac{399}{\frac{119600}{\frac{399}{40400}}}=\frac{101}{299}\)
giúp mình với
Mình nghĩ \(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}\)
\(299A=\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+\frac{299}{3\cdot302}+...+\frac{299}{101\cdot400}\)
\(299A=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\)
\(299A=\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)=C\)
\(A=\frac{C}{299}\)
Lại có;
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+....+\frac{1}{299\cdot400}\)
\(101B=\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+...+\frac{101}{299\cdot400}\)
\(101B=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\)
\(101B=\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)=C\)
\(B=\frac{C}{101}\)
Vậy \(\frac{A}{B}=\frac{C}{299}:\frac{C}{101}=\frac{101}{299}\)