2/1.3 + 2/3.5 + 2/5.7 +...+ 2/97.99 
=(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/97-1/99) 
=1-1/99=98/99 

 

 \(A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

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

\(A=\frac{1}{1}-\frac{1}{99}\)

\(A=\frac{98}{99}\)

ta có A=1-1/3+1/2-1/5+..................1/95-1/97+1/97-1/99

        A=1-1/99

        A=98/99

Cho A =2/1.3+2/3.5+2/5.7+2/7.9+....2/97.99 

A=1-1/3+1/3-1/5+1/5-1/7+..........+1/97-1/98

A=1-1/98

A=98/99

\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)

=3.(3/1.3+3/3.5+3/5.7+...+3/95.97+3/97.99)

=3(1-1/3+1/3-1/5+1/5-1/7+...+1/95-1/97+1/97-1/99)

=3[(1-1/99)+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)]

=3(1-1/99)=3(99/99-1/99)=3.98/99=1.98/33=98/33

Neu la 3 ma ko phai la 3^2 thi sao : Tinh gium minh nha .

A=1.3.(5-2)+3.5.(7-2)+5.7.(9-2)+...+97.99(101-2)=

=(1.3.5+3.5.7+5.7.9+...+97.99.101) - 2.(1.3+3.5+5.7+...+97.99)

Đặt 

B=1.3.5+3.5.7+5.7.9+...+97.99.101

=> 8B=1.3.5.8+3.5.7.8+5.7.9.8+...+97.99.101.8=

=1.3.5(7+1)+3.5.7.(9-1)+5.7.9.(11-3)+...+97.99.101.(103-95)=

=1.3.5+1.3.5.7-1.3.5.7+3.5.7.9-3.5.7.9+5.7.9.11-...-95.97.99.101+97.99.101.103=

=3.5+97.99.101.103

\(\Rightarrow B=\dfrac{15+97.99.101.103}{8}\)

Đặt 

C=1.3+3.5+5.7+...+97.99

=> 6C=1.3.6+3.5.6+5.7.6+...+97.99.6=

=1.3.(5+1)+3.5.(7-1)+5.7.(9-3)+...+97.99.(101-95)=

=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-95.97.99+97.99.101=

=1.3+97.99.101

\(\Rightarrow C=\dfrac{3+97.99.101}{6}\)

\(\Rightarrow A=B-2C\)

Thay các giá trị của B và C bạn tự tính nốt nhé

7x6 hay 7x9 hả bạn

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\)\(\frac{2}{97.99}\)

\(A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.........+\frac{1}{97.99}\right)\)

\(A=2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=2\left(\frac{1}{1}-\frac{1}{99}\right)\)

\(A=2.\frac{98}{99}\)

\(A=\frac{196}{99}\)

Ta có: 2/1.3 = 2/1 - 2/3    ;   2/3.5 = 2/3 - 2/5 ; bạn làm tương tự với các số kia.

Ta được : 2/1 - 2/3 + 2/3 - 2/5 + 2/5 - 2/7 + 2/7 - 2/6 + ....+ 2/97 - 2/99

A = 1 - 2/1 - 2/3 + 2/3 - 2/5 + 2/5 - 2/7 + 2/7 - 2/6 + ....+ 2/97 - 2/99

= 97/99 = > A = 97/99 : 2 = 97/99 x 1/2 = 97/198