Ta có: \(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)

\(=1\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2014.2015}\right)\)

\(=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

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

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

Vậy.....

\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+....+\frac{4}{2014.2015}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2014}-\frac{1}{2015}\)

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

Ta có: 

A = \(\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)

=\(4.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\right)\)

=\(4.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

=\(4.\left(\frac{1}{1}-\frac{1}{2015}\right)\)

=\(4.\frac{2014}{2015}\)

=\(\frac{8056}{2015}\)

\(A=\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)

\(A=4\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\right)\)

\(A=4\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

\(A=4\left(\frac{1}{1}-\frac{1}{2015}\right)\)

\(A=4\left(\frac{2015-1}{2015}\right)\)

\(A=4.\frac{2014}{2015}\)

... BẠN TỰ LÀM NỐT NHÉ!

\(\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+...+\frac{4}{2014\cdot2015}\)

\(=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2014\cdot2015}\right)\)

\(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2014}-\frac{1}{2015}\right)\)

\(=4\left(1-\frac{1}{2015}\right)\)

\(=4\cdot\frac{2014}{2015}=\frac{8056}{2015}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

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

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

A=1/1-1/2+1/2-1/3+1/3-...........+1/2014-1/2015

A=1/1-1/2015

A=2014/2015

=1-1/2+1/2-1/3+1/3-1/4+....+1/2014-1/2015

Trừ tất cả ta được 1-1/2015=2014/2015

=1-1/2+1/2-1/3+1/3-1/4+.....+1/2014-1/2015

=1-1/2015=2014/2015

=1-(1/2+1/2-1/3+1/3-1/4+...+1/2014-1/2015)

=1-1/2015                           

=2014/2015.   

Nếu đúng thì nhớ tíck cho mk nhé!!!Thanh you...

1.

a. \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)

\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=5.\left(1-\frac{1}{100}\right)\)

\(=5.\frac{99}{100}\)

\(=\frac{99}{20}\)

b. \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)

\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

\(=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{4}{2}.\left(1-\frac{1}{101}\right)\)

\(=2.\frac{100}{101}\)

\(=\frac{200}{101}\)

Đặt \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(\Rightarrow\frac{1}{2}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=1-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=\frac{100}{101}\)

\(\Rightarrow A=\frac{100}{101}.2\)

\(\Rightarrow A=\frac{200}{101}.\)

dấu bằng của mk bt liệt nên bạn thông cảm 

A bằng (1.2.3.4).(1.2.3.4)/(1.2.3.4).(2.3.4.5) bằng 5 

rút gọn cho nhau bạn nhé

thank bạn nhóa