#)Thắc mắc ? 

Cho mk hỏi cái ''với 2'' là j bn ? so sánh ak, nếu là so sánh thì mk giải thế này :

#)Giải :

\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)

\(M=2-1+1-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+...+\frac{2}{48}-\frac{2}{49}+\frac{2}{49}-\frac{2}{50}\)

\(M=2-\frac{2}{50}\)

\(M=1\frac{24}{25}=\frac{49}{25}\)

So sánh \(\frac{49}{25}\)với  2

\(2=\frac{2}{1}=\frac{50}{25}\)

Vì \(\frac{49}{25}< \frac{50}{25}\Rightarrow\frac{49}{25}< 2\Rightarrow M< 2\)

          #~Will~be~Pens~#

\(M=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{49.50}=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)=2\left(1-\frac{1}{50}\right)=2x\frac{49}{50}=\frac{49}{25}=1\frac{24}{25}\)

Vì M=1 24/25

=>M<2

Ta có:

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}=\frac{432}{15}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{432}{15}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{432}{15}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{1}-\frac{432}{15}=-\frac{139}{5}\)

\(\Rightarrow\left(x+1\right).-139=5\Rightarrow x=\frac{-5}{139}-1=-\frac{144}{139}\)

Có thể bạn đã sai đề.

1 - 1/x+1 = 432/15

1/x+1 = 1- 432/15 

\(\frac{-139}{5}\)

Ta có\(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{x\left(x+1\right)}=\frac{432}{15}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\)= \(\frac{432}{15}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{432}{15}\)

\(\Leftrightarrow\frac{1}{x+1}=-\frac{139}{5}\)

\(\Leftrightarrow-139\left(x+1\right)=5\)

x=\(\frac{-144}{139}\)

ta có 

\(C=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{4.3}+..+\frac{100-99}{99.100}\)

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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{99}{100}\)

\(=\frac{1}{100}\)

giúp vs

= 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +... + 1/99 - 1/100

= 1 - 99/100

= 1/100.

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=\(\dfrac{1}{1}-\dfrac{1}{100}=\dfrac{100}{100}-\dfrac{1}{100}\)

=\(\dfrac{99}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

= 

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+....\left(\frac{1}{2003}-\frac{1}{2003}\right)-\frac{1}{2004}\)

\(=1-0+0+0+....+0-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

thank you

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

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

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

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

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

= \(\frac{9}{38}\)

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

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

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

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