\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{2010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2010}\)
\(\Leftrightarrow x+1=2010\)
\(\Leftrightarrow x=2009\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{49}{100}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{50}{100}-\frac{2}{100}=\frac{49}{100}\)

~~~~~~~~~~~Ai thấy đúng thì nhanh k nha ~~~~~~

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~ 

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~ 

~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~

Ta có:

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(\Rightarrow\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(A=\frac{1\cdot1}{1\cdot2}\cdot\frac{2\cdot2}{2\cdot3}\cdot\frac{3\cdot3}{3\cdot4}\cdot\frac{4\cdot4}{4\cdot5}=\frac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot3\cdot4}\cdot\frac{1\cdot2\cdot3\cdot4}{2\cdot3\cdot4\cdot5}=\frac{1}{5}\)

A= 1/2 * 2/3 * 3/4 * 4/5

  =  1*2*3*4/2*3*4*5

  =   1/5

A = \(\frac{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5}\)= \(\frac{\left(1\cdot2\cdot3\cdot4\right)\cdot\left(1\cdot2\cdot3\cdot4\right)}{\left(1\cdot2\cdot3\cdot4\right)\cdot\left(2\cdot3\cdot4\cdot5\right)}\)= \(\frac{1}{5}\)

Bạn k cho mik nhé.

E = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

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

E = \(\frac{1}{2}-\frac{1}{100}\)

E = \(\frac{49}{100}\)

\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{100}+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow x-\frac{98}{99}=\frac{1}{99}\Leftrightarrow x=1\)

Mình đang bí đây nè

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

= 1 - 1/100

= 99/100

ta có:1/1\(\times\)2+1/2\(\times\)3+1/3\(\times\)4+...........+1/99\(\times\)100(xin lỗi nha mình không biết viết phân số)

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

=1/1-1/100

=99/100

Ai tihcs mình mình tích lại cho.

Dii vào link nào nha : https://loga.vn/hoi-dap/tinh-tong-s-3-1-2-3-2-3-3-3-4-3-4-5-3-2015-20161-tinh-tong-s-dfrac-3-1-2-dfrac-3-2-3-dfrac-3-3-4-17371 

k mk

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)

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

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)

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

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

\(A=\frac{2005}{2006}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)

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

\(\Rightarrow A=1-\frac{1}{2006}\)

\(\Rightarrow A=\frac{2005}{2006}\)

Vậy \(A=\frac{2005}{2006}\)