Tính tỉ số \(\frac{A}{B}\):
A = \(\frac{1}{1.300}+\frac{1}{2.301}+...+\frac{1}{101.400}\)
và B = \(\frac{1}{1.102}+\frac{1}{1.103}+...+\frac{1}{299.400}\)
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18 tháng 12 2016
\(A=\frac{\frac{1}{1.300}+\frac{1}{2.301}+...+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}}=\frac{1}{154526}\)
HN
30 tháng 9 2021
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11 tháng 5 2016
\(A=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...........+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left(1+\frac{1}{2}+........+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-....-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+........+\frac{1}{299}-\frac{1}{400}\right)\)
\(=\frac{1}{101}.\left(1+\frac{1}{2}+.......+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-............-\frac{1}{400}\right)\)
\(=\frac{1}{101}\left(1+\frac{1}{2}+......+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{299}-\frac{1}{102}-.....-\frac{1}{300}-....-\frac{1}{400}\right)\)
\(=\frac{1}{101}\left(1+\frac{1}{2}+........+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}}{\frac{1}{101}}=\frac{101}{299}\)
11 tháng 5 2016
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
VM
20 tháng 1 2020
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!
12 tháng 5 2015
đặt \(A=\frac{1}{1.300}+\frac{1}{2.301}+...+\frac{1}{101.400}\)
\(\Rightarrow299A=\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)=C\)
\(\Rightarrow A=\frac{C}{299}\)
đặt \(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(\Rightarrow101B=\frac{101}{1.102}+\frac{101}{2.103}+...+\frac{1}{299.400}=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)=\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+...+\frac{1}{400}\right)=C\)
\(\Rightarrow B=\frac{C}{101}\)
bài toán được viết lại như sau:
\(\frac{C}{\frac{299}{\frac{C}{101}}}\)=\(\frac{101}{299}\)
Sửa đề: \(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+...+\frac{1}{299\cdot400}\)
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\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+...+\frac{1}{101\cdot400}\\ A=\frac{1}{299}\left(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+...+\frac{299}{101\cdot400}\right)\\ A=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\\ A=\frac{1}{299}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+\frac{1}{302}+...+\frac{1}{400}\right)\right]\left(1\right)\)
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+...+\frac{1}{299\cdot400}\\ B=\frac{1}{101}\left(\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+...+\frac{101}{299\cdot400}\right)\\ B=\frac{1}{101}\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\\ B=\frac{1}{101}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)\right]\\ B=\frac{1}{101}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}{\frac{1}{101}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}\\ \frac{A}{B}=\frac{\frac{1}{299}}{\frac{1}{101}}=\frac{1}{299}\cdot101=\frac{101}{299}\)
Chúc bạn học tốt!