Tính \(\dfrac{A}{B}\) biết:
A = \(\dfrac{1}{1.300}\) + \(\dfrac{1}{2.301}+\dfrac{1}{3.302}+.....+\dfrac{1}{101.400}\)
B = \(\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+.....+\dfrac{1}{299.400}\)
K
Khách
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
20 tháng 9 2021
Lời giải:
\(299A=\frac{300-1}{1.300}+\frac{301-2}{2.301}+\frac{302-3}{3.302}+....+\frac{400-101}{101.400}\)
\(=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)
\(=(1+\frac{1}{2}+....+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(1)\)
Mặt khác:
$101B=\frac{102-1}{1.102}+\frac{103-2}{2.103}+...+\frac{400-299}{299.400}$
$=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}$
$=(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{299})-(\frac{1}{102}+\frac{1}{103}+....+\frac{1}{400})$
$=(1+\frac{1}{2}+...+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(2)$
Từ $(1);(2)\Rightarrow 299A=101B$
$\Rightarrow \frac{A}{B}=\frac{101}{299}$
11 tháng 4 2017
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 ~
DT
Dang Tung
CTVHS
25 tháng 6 2023
`a)` Xét tử số phân số M :
\(2012-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{2012}{2020}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{2012}{2020}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{2020}\\ =24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)\)
Ta được : \(M=\dfrac{24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)}{\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}}=24\)
DT
Dang Tung
CTVHS
25 tháng 6 2023
`b)` Xét tử số phân số N :
\(\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+...+\dfrac{1}{101.400}\\ =\dfrac{1}{299}.\left(\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+...+\dfrac{299}{101.400}\right)\\ =\dfrac{1}{299}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)
Xét mẫu số phân số N :
\(\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+...+\dfrac{1}{299.400}\\ =\dfrac{1}{101}.\left(\dfrac{101}{1.102}+\dfrac{101}{2.103}+\dfrac{101}{3.104}+...+\dfrac{101}{299.400}\right)\\ =\dfrac{1}{101}.\left(1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+\dfrac{1}{3}-\dfrac{1}{104}+...+\dfrac{1}{299}-\dfrac{1}{400}\right)\)
\(=\dfrac{1}{101}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)
Ta được: \(N=\dfrac{\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}{\dfrac{1}{101}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}\\ =\dfrac{\dfrac{1}{299}}{\dfrac{1}{101}}=\dfrac{101}{299}\)
DP
25 tháng 6 2017
\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{3012}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\frac{399}{400}\)
\(A=\frac{399}{119600}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\frac{399}{400}\)
\(B=\frac{399}{40400}\)
\(\Rightarrow\frac{A}{B}=\frac{399}{\frac{119600}{\frac{399}{40400}}}=\frac{101}{299}\)
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
NT
1
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
A/B =1
Không chắc vì tính nhẩm
Kết quả là \(\dfrac{101}{299}\). Cô mình chữa rồi đó, và mình lúc đầu cũng không làm được.