\(\frac{2004.2006+1000}{2005.2006-1006}\)

\(=\frac{2004.2006+1000}{\left(2004+1\right).2006-1006}\)

\(=\frac{2004.2006+1000}{2004.2006+2006-1006}\)

\(=\frac{2004.2006+1000}{2004.2006+1000}\)

\(=1\)

\(\frac{2004x2006+1000}{2005x2006-1006}\)

=\(\frac{1000}{1x2006-1006}\)cùng giảm 2004 lần số 2006

=\(\frac{1000}{1000}=1\)

\(\frac{2005.2007-1}{2004+2005.2006}=\frac{2005.2006+2005-1}{2004+2005.2006}=\frac{2004+2005.2006}{2004+2005.2006}=1\)

\(A=\frac{1}{500}+\frac{3}{500}+\frac{5}{500}+...+\frac{95}{500}+\frac{97}{500}+\frac{99}{500}\)

\(A=\frac{1+3+5+...+95+97+99}{500}\)

\(A=\frac{\left(1+99\right)x50:2}{500}=\frac{100x50:2}{500}=\frac{100x5x10x\frac{1}{2}}{100x5}=10x\frac{1}{2}=5\)

\(\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}}{500-\frac{500}{501}-\frac{501}{502}-...-\frac{999}{1000}}=\frac{\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{999}-\frac{1}{1000}\right)}{500-\left(1-\frac{1}{501}\right)-\left(1-\frac{1}{502}\right)-...-\left(1-\frac{1}{1000}\right)}\)

hình như cái mẫu bạn ghi dấu sai thì phải, còn tử thì mình lười làm lắm

tử bạn tính ra 1/2+1/12+...+1/999 000 sau đó phân tích ra là

khó thật

nhớ L-I-K-E nhe tại vì cậu bảo giúp mình, mình cho đúng liền

A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{2005}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
= \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
= \(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\)\(-\frac{1}{1}-\frac{1}{2}-...-\frac{1}{1003}\)
= \(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2005}+\frac{1}{2006}\)
(=) B - A = \(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}+\frac{1}{2016}\)- \(\frac{1}{1004}-\frac{1}{1005}-...-\frac{1}{2005}-\frac{1}{2006}\)
= \(\frac{1}{2007}+\frac{1}{2008}+...+\frac{1}{2016}-\) \(\frac{1}{1004}-\frac{1}{1005}-\frac{1}{1006}-\frac{1}{1007}\)

Bó tay, sai đề rồi bn à, nếu tính đc thì cũng dài dòng lắm...........

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)\)

\(=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)(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:

\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}\)