bài 1 :
A = 1/ 1.2 + 1/3.4 + 1/5.6 + .........+ 1/ 2005 . 2006
B = 1/ 1004.2006 + 1/ 1005.2006 + ......+ 1/2006.1004
CMR: A/B thuộc Z ( số nguyên )
Những câu hỏi liên quan
A=1/1.2+1/3.4+...+1/2005.2006 và B=1/1004.2006+1/1005.2006+...+1/2006.1004
CMR A/B thuộc Z
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005.2006}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(A=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)\)
\(A=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.1004}\)
\(3010B=\frac{1004+2006}{1004.2006}+\frac{1005+2005}{1005.2005}+...+\frac{2006+1004}{2006.1004}\) ( sửa đề nhé )
\(3010B=\frac{1}{2006}+\frac{1}{1004}+\frac{1}{2005}+\frac{1}{1005}+...+\frac{1}{1004}+\frac{1}{2006}\)
\(3010B=2\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)
\(B=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)
\(\Rightarrow\)\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}=1505\) hay \(\frac{A}{B}\inℤ\)
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Đặt A=1/1.2+1/3.4+...+1/2005.2006,B=1/1004.2006+1/1005.2006+...+1/2006.1004 Chứng minh rằng A/B thuộc Z
Cho A=1.1/2+1.3/4+1./5.6+....+1.2005/2006
B= 1/1004.2006+1/1005.2006+...+1/2006.1004
Chứng minh rằng A/B là môt số nguyên
Giúp mk na
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\);\(B=\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.1004}\)
Chứng minh rằng \(\frac{A}{B}\)thuộc Z
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{1004.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.1004}\)
Tính \(\frac{A}{B}\)
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
A=1/1.2+1/3.4+...+1/2005.2006
B=1/1004.2006+1/1005.12005+...+1/2006.1004
Chung minh A/B la so nguyen
Đặt A=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)
B=\(\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.1004}\)
Chứng tỏ rằng \(\frac{A}{B}\in Z\)
Ai làm được nhanh và đúng tớ tick đúng nhé
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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