Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+................+\dfrac{2}{50.52}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..............+\dfrac{1}{50}-\dfrac{1}{52}\)
\(=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+.....+\dfrac{2}{50.52}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+.....+\dfrac{1}{50}-\dfrac{1}{52}\)
\(=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{52}=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x.\left(x+2\right)}=\dfrac{4}{9}\)
=\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}\)
=\(\dfrac{1}{2}-\dfrac{1}{x+2}\)=\(\dfrac{4}{9}\)
=>\(\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}=\dfrac{1}{18}\)
=>\(x+2=18\)
=>x=16
Đặt A = 2/2.4+2/4.6+....+2/2010.2012
= 1/2 - 1/4 + 1/4 - 1/6 + ...... + 1/2010 - 1/2012
= 1/2 - 1/2012
= 1005/2012
2/2.4+2/4.6+...+2/2010.2012
= 1/2-1/4+1/4-1/6+...+1/2010-1/2012
= 1/2-1/2012
= 1006/2012-1/2012
= 1005/2012