x + 25 = 64

x         = 64 - 25

x         = 39

Vậy x = 39

A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

= \(1-\frac{1}{50}=\frac{49}{50}\)

B = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

= \(2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\right)\)

= \(2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)

= \(\frac{2}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)

= \(\frac{4}{13}\)

C = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)

= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)

= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\) 

= \(\frac{9}{38}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)

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

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

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{37.39}\)

\(=\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{39-37}{37.39}\)

\(=\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+...+\frac{39}{37.39}-\frac{37}{37.39}\)

\(=\)\(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{37}-\frac{2}{39}\)

\(=\frac{2}{3}-\frac{2}{39}\)

\(=\frac{8}{13}\)

Ta có:  2/3.5=1/3 - 1/5

           2/5.7=1/5 - 1/7

                  ........

           2/37.39=1/37 - 1/39

2/3.5 + 2/5.7 + ... + 2/37.39=1/3 - 1/5 + 1/5 - 1/7 +..... + 1/37 - 1/39 

                                        = 1/3 - 1/39

                                        = 36/117

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2017\cdot2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b) \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)( sửa 91.99 thành 97.99 mới đúng nha )

\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{99}\right)\)

\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{99}\right)\)

\(=\frac{1}{2}.\frac{64}{99}\)

\(=\frac{32}{99}\)

a) 1/1.2 + 1/2.3 + 1/3.4 +...+1/2017.2018

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ....+1/2017 - 1/2018

= 1 - 1/2018 

= 2017/2018

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

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

\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{49}{50}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}\)

A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

= \(\left(1-\dfrac{1}{2}\right)\)+\(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\)+...+\(\left(\dfrac{1}{49}-\dfrac{1}{50}\right)\)

= \(\left(1-\dfrac{1}{50}\right)\) = \(\dfrac{49}{50}\)

s = 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5

S=1 + (-1/2 +1/2)+...+(-1/4 + 1/4 ) +-1/5

S = 1 + 0 +0 +...+ 0 +-1/5

S= 1 + -1/5

S = 4/5

S=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5

S=1-1/5

S=4/5.

P=1/1.3+1/3.5+1/5.7+1/7.9

2P=2/1.3+2/3.5+2/5.7+2/7.9

2P=1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9

2P=1-1/9=8/9

P=8/9:2

P=4/9.

Chac chan dung nha ban.k cho minh nhe

S=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5 =1/1-1/5 nhé  P=1/2+(2/1.3+2/3.5+2/5.7+2/7.9)=1/2+(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9).=1/2+(1/1-1/9)

Ta có: \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2005.2006}\)

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

          \(=1-\frac{1}{2006}=\frac{2005}{2006}\)

 \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2015.2017}\)

      \(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}\)

        \(=1-\frac{1}{2017}=\frac{2016}{2017}\)

N = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/2005 - 1/2006

   = 1/1 - 1/2006

   = 2006/2006 - 1/2006

   =  2005/2006

1)

2/3.5+2/5.7+...+2/11.13+2/13.15+2/1.2+2/2.3+...+2/9.10

=(2/3.5+...2/13.15)+(2/1.2+...+2/9.10)

= (2/3-2/15)+ [2(1-1/10)]

=8/15+9/5

=7/3

2)

11/12+11/12.24+...+11/88.99

=11-1/9

=10/8/9