Tính tỉ số \(\frac{A}{B}\)biết:
\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
và \(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
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Tính: \(\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}}\)
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\(\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+.............+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+.......+\frac{1}{299.400}}\)
20 tháng 1 2020 lúc 10:29
tính tỉ số\(\frac{A}{B}\)bằng cách nhanh nhất biết :
A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
B=\(\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
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!
Tính:
A=\(\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}}\)
Giúp mình nha các bạn
cho \(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+.....+\frac{1}{101.400}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+....+\frac{1}{299.400}\)
so sánh A và B
Tính: \(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.4000}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}}\)
Cho \(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
Tính \(A:B\)
Tính bằng cách hợp lí:
a) A=\(\left(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\right):\left(\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\right)\)
b) B=\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{200}\right):\left(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....\frac{198}{2}+\frac{199}{1}\right)\)
S= 1/199 + 2/198 + ... + 198/2 + 199/1
S= (1/199 + 1) + (2/198 + 1)+ ... + (198/2 + 1) +1
S= 200/200 + 200/199 + 200/198 + ... + 200/2
S= 200.(1/200 + 1/199 + ... + 1/2)
Suy ra , B=(1/2 + 1/3 + ... +1/200) : 200.(1/2 + 1/3 + ... + 1/200)
B=1 : 200 = 1/200
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Tính A=\(\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+....+\frac{1}{299.400}}\)
\(A=\frac{\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}}{\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}}\)
\(A=\frac{\frac{1}{299}\left(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+\frac{299}{3\cdot302}+...+\frac{299}{101\cdot400}\right)}{\frac{1}{101}\left(\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+\frac{101}{3\cdot104}+...+\frac{299}{299\cdot400}\right)}\)
\(A=\frac{\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\right)}{\frac{1}{101}\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+\frac{1}{3}-\frac{1}{104}+...+\frac{1}{299}-\frac{1}{400}\right)}\)
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