\(\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)\): \(\left(\frac{2}{26}+\frac{2}{27}+...+\frac{2}{50}\right)\)
Những câu hỏi liên quan
CHỨNG MINH RẰNG:
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
Ta có vế trái:
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}=\)vế phải
Vậy\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\RightarrowĐpcm\)
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CMR
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Chứng minh rằng :
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Ta có:
1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50
= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/49 - 1/50
= (1 + 1/3 + 1/5 + .... + 1/49) - (1/2 + 1/4 + 1/6 + .... + 1/50)
= (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/49 + 1/50) - 2.(1/2 + 1/4 + 1/6 + ... + 1/50)
= (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/49 + 1/50) - (1 + 1/2 + 1/3 + ... + 1/25)
= 1/26 + 1/27 + 1/28 + ... + 1/50
=> đpcm
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Chứng tỏ rằng:
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+.......+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+........+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.......+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+........+\frac{1}{50}\left(đpcm\right)\)
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Chứng minh rằng:
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}.\)
giải cụ thể ra giùm.
Ta có
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{2-1}{1.2}+\frac{3-2}{3.4}+...+\frac{50-49}{49.50}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+...+\frac{50}{49.50}-\frac{49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(1-\frac{1}{50}=\frac{49}{50}\)
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Tính:Afrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+frac{1}{3cdot4cdot5}+...+frac{1}{48cdot49cdot50}Bleft(1-frac{1}{2}right)+left(1-frac{1}{4}right)+left(1-frac{1}{8}right)+...+left(1-frac{1}{1024}right)C4cdot5^{100}left(frac{1}{5}+frac{1}{5^2}+frac{1}{5^3}+...+frac{1}{5^{100}}right)D1+frac{9}{45}+frac{9}{105}+frac{9}{189}+frac{9}{29997}Không cần làm hết cũng đc, giúp tớ nha
Đọc tiếp
Tính:
A=\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{48\cdot49\cdot50}\)
B=\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+...+\left(1-\frac{1}{1024}\right)\)
C=\(4\cdot5^{100}\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\right)\)
D=\(1+\frac{9}{45}+\frac{9}{105}+\frac{9}{189}+\frac{9}{29997}\)
Không cần làm hết cũng đc, giúp tớ nha
bạn tách ra xong làm cx dễ mà đây là toán 6
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Cảm ơn câu trả lời thật súc tích và thật ngắn gọn của bạn
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\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)
\(A=\frac{1}{2}.\frac{612}{1225}=\frac{306}{1225}\)
~ Hok tốt ~
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Bài 1:
a) \(\frac{1}{1}\cdot2+\frac{1}{2}\cdot3+\frac{1}{3}\cdot4+...+\frac{1}{n}\cdot\left(n+1\right)\)
b) \(\frac{1}{1}\cdot2\cdot3+\frac{1}{2}\cdot3\cdot4+\frac{1}{3}\cdot4\cdot5+...+\frac{1}{a}\cdot\left(a+1\right)\cdot\left(a+2\right)\)
Tính tổng :a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
Tính tổng :
a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}
b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}
c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}
d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}
e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
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