a)1/5.6+1/6.7+1/7.8+.......+1/99.100

= (1/5-1/6)+(1/6-1/7)+(1/7-1/8)+.....+(1/99-1/100)

= 1/5 - 1/100

= 19/100

b)2/1.3+2/3.5+2/5.7+.........+2/2013.2015

= (1/1-1/3)+(1/3-1/5)+(1/5-1/7)+.....+(1/2013+1/2015)

= 1/1 - 1/2015

= 2014/2015

\(a,\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{5}-\frac{1}{100}=\frac{20}{100}-\frac{1}{100}=\frac{19}{100}\)

\(b,\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)

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

\(=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2013.2015}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2013.2015}\right)\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)

Vậy A=1007/2015

\(2A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(2A=1-\frac{1}{2015}\)

\(A=\frac{2014}{2015}:2\)

\(A=\frac{1007}{2015}\)

1/1.3+1/3.5+...+1/2013.2015

=1/2.(1/1-1/3+1/3-1/5+...+1/2013-1/2015)

=1/2.(1/1-1/2015)

=1/2.2014/2015

=1007/2015

A=1/1.3+1/3.5+1/5.7+...+1/2013.2015

2A=2.(1/1.3+1/3.5+1/5.7+...+1/2013.2015)

=2/1.3+2/3.5+2/5.7+...+2/2013.2015

=1-1/3+1/5-1/7+1/7-1/9+...+1/2013-1/2015

=1-1/2015

=2014/2015

=>2A=2014/2015=>A=1007/2015

A=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

A=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\) (phân số 1/2 là bn xem hai số của mẫu,số đuôi trừ số đầu có ra kết quả bằng tử hay ko,nếu ko bằng,bạn phải nhân thêm với 1 phân số có tử là tử của các phân số cho sẵn còn mẫu là hiệu của hai số ở mẫu)

A=\(\frac{1}{2}.\left(1-\frac{1}{2015}\right)\) (những phân số giống nhau thì cứ loại bỏ)

A=\(\frac{1}{2}.\frac{2014}{2015}=1.\frac{1007}{2015}=\frac{1007}{2015}\)

\(B=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{2013\times2015}\\ 2B=\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{2013\times2015}\\ 2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\\2B=1-\frac{1}{2015}=\frac{2014}{2015}\\ \Rightarrow B=\frac{2014}{2015} \div2=\frac{1007}{2015}\)

\(M=1-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)\)

\(M=1-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)\)

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

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

bạn tự tính nốt nhé

\(M=1-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-...-\frac{1}{2013.2015}\)

\(\Leftrightarrow M=1-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\right)\)

\(\Leftrightarrow M=1-\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

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

\(\Leftrightarrow M=1-\frac{1}{2}\left(1-\frac{1}{2015}\right)\)

\(\Leftrightarrow M=1-\frac{1}{2}.\frac{2014}{2015}\)

\(\Leftrightarrow M=1-\frac{2014}{4030}\)

\(\Leftrightarrow M=\frac{2016}{4030}=\frac{1008}{2015}\)

Ta có: A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2013.2015}\)

\(\Leftrightarrow2A=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2013.2015}\right)\)

\(\Leftrightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2013}+\dfrac{1}{2013}-\dfrac{1}{2015}\)

\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{2015}=\dfrac{2012}{6045}\)

\(\Leftrightarrow A=\dfrac{1006}{6045}\)

2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{2013.2015}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}+\dfrac{1}{2015}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{2015}\)

2A=\(\dfrac{2014}{2015}\)

 A=\(\dfrac{1007}{2015}\)

                     Khi gặp bài này, bn nên tách 1 phân số ra thành hiệu của 2 phân số.

Giải:

A=1/1.3+1/3.5+1/5.7+...+1/2013.2015

A=1/2.(2/1.3+2/3.5+2/5.7+...+2/2013.2015)

A=1/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/2013-1/2015)

A=1/2.(1/1-1/2015)

A=1/2.2014/2015

A=1007/2015

Chúc bạn học tốt!

= 1/2. ( 1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +........+ 1/2013 - 1/2015)

= 1/2 . ( 1- 1/2015)

= 1007/2015

hỏi gì nhiều thế