A =1/1*300 + 1/2*301 + 1/302 + ... + 1/101 * 400
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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
Tính A/B biết rằng:
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
1/1×300 + 1/2×301 + 1/3×302 + ..... + 1/101×400
1/1×102+1/2×103+1/3×104 + ........+ 1/299×400
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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
tính A:B bt A=1/1*300+1/2*301+...+101*400
B=1/1*102+11/2*103+...+1/299*400
tính A/B biết:
A=1/1*300+1/2*301+....+1/101*400
B=1/1*102+1/2*103+...+1/299*400
29 tháng 8 2021 lúc 15:28
Tính A/B biết rằng: A= 1/1*300 + 1/2*301 +...+ 1/101*400 B= 1/1*102 + 1/2*103 +...+ 1/298*399 + 1/299*400
\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+...+\dfrac{1}{101.400}\)
\(\Rightarrow299A=\dfrac{299}{1.300}+\dfrac{299}{2.301}+...+\dfrac{299}{101.400}=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+...+\dfrac{1}{101}-\dfrac{1}{400}=M\)
\(\Rightarrow A=\dfrac{M}{299}\left(1\right)\)
Ta lại có:
\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+...+\dfrac{1}{298.399}+\dfrac{1}{299.400}\)
\(\Rightarrow101B=\dfrac{101}{1.102}+\dfrac{101}{2.103}+...+\dfrac{101}{399.400}=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+...+\dfrac{1}{399}-\dfrac{1}{400}=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+...+\dfrac{1}{101}-\dfrac{1}{400}=M\)
\(\Rightarrow B=\dfrac{M}{101}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{A}{B}=\dfrac{M}{299}:\dfrac{M}{101}=\dfrac{101}{299}\)
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tìm A/B biết cứu với hix hix
A= 1/(1*300)+1/(2*301)+....+1/(101*400)
B=1/(1*102)+1/(1/2*103)+....+1/(499*400)
\(\frac{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+\frac{1}{302}+...+\frac{1}{400}\right)}{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+\frac{1}{104}+...+\frac{1}{400}\right)}\)
G mk với, mk cần gấp lắm. Ai giải được mk k cho