Cho A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+......+\frac{1}{2013.2014}\)
B=\(\frac{1}{1008.2014}+\frac{1}{1009.2013}+\frac{1}{1010.2012}+......+\frac{1}{2014.1008}\)
Chứng tỏ rằng:\(\frac{A}{B}\) là số nguyên
Những câu hỏi liên quan
CHO A= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2013.2014}\\ \)
B=\(\frac{1}{1008.2014}+\frac{1}{1009.2013}+\frac{1}{1010.2012}+...+\frac{1}{2014.1008}\)
Hãy Chứng tỏ rằng \(\frac{A}{B}\)LÀ SỐ NGUYÊN
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2013.2014}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2013}+\frac{1}{2014}-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2013}+\frac{1}{2014}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\)
Lại có B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+\frac{1}{1010.2012}+...+\frac{1}{2014.1008}\)
=> 3022B = \(\frac{3022}{1008.2014}+\frac{3022}{1009.2013}+\frac{3022}{1010.2012}+...+\frac{3022}{2014.1008}\)
\(=\frac{1}{1008}+\frac{1}{2014}+\frac{1}{1009}+\frac{1}{2013}+\frac{1}{1010}+\frac{1}{2012}+...+\frac{1}{2014}+\frac{1}{1008}\)
\(=2.\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
=> \(B=\frac{1}{1511}.\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)}{\frac{1}{1511}.\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)}=\frac{1}{\frac{1}{1511}}=1511\)
=> \(\frac{A}{B}=1511\)
=> A/B là 1 số nguyên (đpcm)
a) Chứng tỏ rằng
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) Đặt A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\); B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng \(\frac{A}{B}\)là số nguyên
Cho A= \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
B= \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng: \(\frac{A}{B}\) là số nguyên
Câu hỏi thì không trả lời mà toàn zô đây bình luận. Haizzz....Thế này phải đổi HOC24 thành CHAT24 thì đúng hơn đó 
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Bình luận (4)
Cho:
A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+......+\frac{1}{2013.2014}\)
B=\(\frac{1}{1008.2014}+\frac{1}{1009.2013}+........+\frac{1}{2014.1008}\)
Chứng tỏ \(\frac{A}{B}\) là số nguyên.
a) Chứng tỏ rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) Đặt A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013+2014}\); Đặt B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng \(\frac{A}{B}\)là số nguyên
cho A frac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+...+frac{1}{2013.2014}Bfrac{1}{1008.2014}+frac{1}{1009.2003}+frac{1}{1010.2012}+...+frac{1}{2014.1008}chứng tỏ frac{A}{B}là số nguyên
Đọc tiếp
cho A =\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{2013.2014}\)
B=\(\frac{1}{1008.2014}\)+\(\frac{1}{1009.2003}\)+\(\frac{1}{1010.2012}\)+...+\(\frac{1}{2014.1008}\)
chứng tỏ \(\frac{A}{B}\)là số nguyên
Cho A=1/1.2+1/3.4+1/5.6+...+1/2013.2014
B=1/1008.2014+1/1009.2013+1/1010.2012+...+1/2014.1008
Hãy chứng tỏ A/B là số nguyên
Câu 1
Cho A= (1+2012/1)(1+2012/2)...(1+2012/1000)
B=(1+1000/1)(1+1000/2).....(1+1000/2012)
Tính \(\frac{A}{B}\)
Câu 2
cho E = 1/1.2 + 1/3.4 +1/5.6 +....+1/2013.2014
F= 1/1008.2014 + 1/1009,2013 +....+1/2014.1008
Tính \(\frac{E}{F}\)
Chứng minh rằng:
a)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
b)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< 1-\frac{1}{2.3}\)
Cần gấp, ai nhanh mik tick nha
Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
