1, Cho
A = \(\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{2005.2006}\)
B = \(\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+.......+\frac{1}{2006.1004}\)
Hãy Tính \(\frac{A}{B}\)
Những câu hỏi liên quan
Cho:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+.....+\frac{1}{2003.2004}+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+.....+\frac{1}{2006.1004}\)
Hãy tính \(\frac{A}{B}\)
Cho
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2003.2004}+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+...+\frac{1}{2006.1004}\)
Tìm \(\frac{A}{B}\)
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\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2003.2004}+\frac{1}{2005.2006}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)\(=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+...+\frac{1}{2006.1004}\)
=>3010B=\(\frac{1}{1004}+\frac{1}{2006}+\frac{1}{1005}+\frac{1}{2005}+...+\frac{1}{2006}+\frac{1}{1004}=2\cdot\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)
=>B=\(\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)
=>\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}=1505\)
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Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+............+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+.....+\frac{1}{2006.1004}\)
Tính \(\frac{A}{B}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)\)
\(=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)(1)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+....+\frac{1}{2006.1004}\)
\(\Rightarrow\frac{1}{1004}+\frac{1}{2006}+\frac{1}{1005}+\frac{1}{2005}+...+\frac{1}{2006}+\frac{1}{1004}=2\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)
\(=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)(2)
Thế (1) và (2) vào ta có:
\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}\)
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\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2003.2004}+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+...+\frac{1}{2006.1004}\)
Tính \(\frac{A}{B}\)
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\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\);\(B=\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.1004}\)
Chứng minh rằng \(\frac{A}{B}\)thuộc Z
Bài 1 : Tính hợp lý :a) A 182 . left(frac{1+frac{1}{3}+frac{1}{9}+frac{1}{27}}{2+frac{2}{3}+frac{2}{9}+frac{2}{27}}:frac{4-frac{4}{7}+frac{4}{49}-frac{4}{343}}{1-frac{1}{7}+frac{1}{49}-frac{1}{343}}right) : frac{919191}{808080}b) C -2+frac{1}{-2+frac{1}{-2+frac{1}{-2+frac{1}{3}}}}c) D left(frac{1}{1.2}+frac{1}{3.4}+...+frac{1}{2005.2006}right): left(frac{1}{1004.2006}+frac{1}{1005.2005}+...+frac{1}{2006.1004}right)
Đọc tiếp
Bài 1 : Tính hợp lý :
a) A = 182 . \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\) : \(\frac{919191}{808080}\)
b) C = \(-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{3}}}}\)
c) D = \(\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\right)\): \(\left(\frac{1}{1004.2006}+\frac{1}{1005.2005}+...+\frac{1}{2006.1004}\right)\)
Cho A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{2005.2006}\)
B=\(\frac{1}{1004.2006}\)+\(\frac{1}{1005.2005}\)+...+\(\frac{1}{2006.1004}\)
Chứng minh \(\frac{A}{B}\) là số nguyên
Cho A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)
B=\(\frac{1}{1004.2006}+\frac{1}{1005.2006}+\frac{1}{1006.2006}+...+\frac{1}{2006.2006}\)
Tính A chia B
Tìm x biết
x. (1/1.2 + 1/3.4+ 1/5.6+ ....+ 1/2005.2006) = 1/1004.2006+1/1005.2005+1/1006.2004+ .....+ 1/2006.1004