tính a /b biết
A= 1/1.300+1/2.301+1/3.2012+..+1/101.400
B=1/1.102+1/2.103+1/3.104+...+1/299.400
Những câu hỏi liên quan
tính a /b biết
A= 1/1.300+1/2.301+1/3.2012+..+1/101.400
B=1/1.102+1/2.103+1/3.104+...+1/299.400
Tính tỉ số A/B biết:
A=1/1.300+1/2.301+1/3.302+...+1/101.400 và
B=1/1.102+1/2.103+1/3.104+...+1/299.400
tính A/B biết rằng
A= 1/(1.300)+1/(2.301)+...+1/(101.400)
B=1/(1.102)+1/(2.103)+1/(3.104)+...+1/(299.400)
Tính A/B biết :
A = 1/1.300 + 1/2.301 + 1/3.302 + ... + 1/101.400
B = 1/1.102 + 1/2.103 + 1/3.104 + ...+ 1/299.400
\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{3012}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\frac{399}{400}\)
\(A=\frac{399}{119600}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\frac{399}{400}\)
\(B=\frac{399}{40400}\)
\(\Rightarrow\frac{A}{B}=\frac{399}{\frac{119600}{\frac{399}{40400}}}=\frac{101}{299}\)
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Tính A/B, biết rằng:
A=1/1.300+1/2.301+1/3.302+...+1/101.404
B=1/1.102+1/2.103+1/3.104+..+1/299.400
Tính A/B, biết rằng:
A=1/1.300+1/2.301+1/3.302+...+1/101.404
B=1/1.102+1/2.103+1/3.104+..+1/299.400
Tính A/B :
A= 1/1.300+1/2.301+1/3.302+...+1/101.400
B= 1/1.102+1/2.103+1/3.104+...+1/299.400
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
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Tính tỉ số \(\dfrac{A}{B}\) biết:
\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+...+\dfrac{1}{101.400}\) và
\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+...+\dfrac{1}{299.400}\)
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 $(1);(2)\Rightarrow 299A=101B$
$\Rightarrow \frac{A}{B}=\frac{101}{299}$
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{1/1.300+1/2.301+1/3.302+...+1/101.400}:{1/1.102+1/2.103+1/3.104+...=1/299.400}
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
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