a/

S = 1-2+3-4+5-6+...+2001-2002+2003

   = [-1] +[-1] +...+[-1] +2003

      ------------------------

       1001 số -1

= -1001 +2003 = 1002

b/

A = \(6.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)=6.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)=6.\left(\frac{1}{3}-\frac{1}{2015}\right)=\frac{6.2012}{6045}=\frac{4024}{2015}\)

Mk giải ko chép lại đề nhá!

Bài 3: 

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}\)\(-\frac{1}{50}\)

\(=\frac{1}{1}-\frac{1}{50}\)

\(=\frac{50}{50}-\frac{1}{50}\)

\(=\frac{49}{50}\)

Vậy: M < 1

Bài 2:

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(=\frac{1}{1}-\frac{1}{2015}\)

\(=\frac{2015}{2015}-\frac{1}{2015}\)

\(=\frac{2014}{2015}\)

B1

A co (100-1)+1=100 so hang

A=(100+1).100 :2=5050

a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)

\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)

\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)

Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)

\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)

\(2A=\frac{12}{3}-\frac{12}{99}\)

\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)

1.5/1-5/4+5/4-5/7+5/7-5/9 +....+5/27-5/30

=5/1-5/30

=145/30=29/6.

B = \(\frac{2^3.5.7.5^2.7^3}{\left(2.5.7^2\right)^2}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2.5.1}{1.1.1}=10\)

C = \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\right)\)\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}\left(\frac{33}{99}-\frac{1}{99}\right)=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)

1) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

Đi học về rồi giải cho , giờ đi học đã 

Trong câu hỏi tương tự mà có mới hay !

ko biết lướt hộ cái nhức cả đầu -_-

Q = \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

Q = \(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{2013.2015}\right)\)

Q =  \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2015}\right)\)

Q = \(\frac{1}{2}.\frac{2012}{6045}=\frac{1002}{6045}\)

\(Q=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2013.2015}\)

\(\Rightarrow Q.2=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2013.2015}\right)\)

\(\Rightarrow Q.2=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2013.2015}\)

\(\Rightarrow Q.2=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(\Rightarrow Q.2=\frac{1}{3}-\frac{1}{2015}\)

\(\Rightarrow Q.2=\frac{2012}{6045}\)

\(\Rightarrow Q=\frac{2012}{6045}.\frac{1}{2}=\frac{1006}{6045}\)

Mk tinh nhẩm, nên ko bt kết quả có đúng ko

nên bn thử tính lại kết quả nha!!!

Chúc bn hok tốt!!!

có công thư rồi mà bài này dễ đợi mk 3' nhé mk giải cho

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2012}{6045}=\frac{1006}{6045}\)

lm tắt cu~g chả bt đúng ko ^^, thông cảm 

\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2015}\right)\)

\(\frac{1}{2}.\frac{2012}{6045}\)

=\(\frac{1006}{6045}\)  

ok bạn nhé