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

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

\(A=2.\left(\frac{1}{2}-\frac{1}{100}\right)=2.\frac{49}{100}=\frac{49}{50}\)

Cảm ơn bạn nhiều nha Nguyễn Tuấn Minh

Trả lời

a) \(\frac{-2}{2\cdot3}+\frac{-2}{3\cdot4}+\frac{-2}{4\cdot5}+...+\frac{-2}{19\cdot20}\)

\(=-2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}\right)\)

\(=-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(=-2\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=-2\cdot\frac{9}{20}\)

\(=\frac{-18}{20}=\frac{-9}{10}\)

ĐKXĐ: \(x\ne0;x\ne-1\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2008}{2010}\)

\(\Leftrightarrow2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2008}{2010}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)(Biết công thức này chứ?)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)

\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2008}{2010}\)

\(\Leftrightarrow\dfrac{x-1}{x+1}=\dfrac{2008}{2010}\Leftrightarrow2010x-2010=2008x+2008\Leftrightarrow x=2009\left(tm\right)\)

Vậy x = 2009

Ta có : 1.98 + 2.97 + 3.96 + ...+ 98.1 = 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + .....+ ( 1 + 2 + 3 + ...+ 97 + 98 ) = \(\frac{1.2}{2}\)+ \(\frac{2.3}{2}\)+  \(\frac{3.4}{2}\)+ ...+ \(\frac{98.99}{2}\)= \(\frac{1}{2}\)( 1 . 2 + 2 . 3 + 3 . 4 +...+ 98 . 99).

Vậy A = \(\frac{1}{2}\)

Nè bạn giải cụ thể chi tiết cho mình đk k thì mình mới k cho đk

đặt A = 1.2. + 2.3 + 3.4 + ... + 49.50

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

3A = 49.50.51

A = 41650

Thay vào ta được

41650 + 1/2x = 40642

=> 1/2x = 1008

=> x = 2016