\(2S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{23+24+25}=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+\left(\dfrac{1}{23.24}-\dfrac{1}{24.25}\right)\)\(=\dfrac{1}{1.2}-\dfrac{1}{24.25}=\dfrac{299}{600}\)
Vậy \(S=\dfrac{299}{600}\div2=\dfrac{299}{1200}\)
(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+...+\(\dfrac{1}{8.9.10}\)).x=\(\dfrac{23}{45}\)
Tính A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{37.38.39}\)
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)
\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+....+\(\dfrac{1}{8.9.10}\)x=\(\dfrac{44}{45}\)
Tính giá trị biểu thức:
\(C=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{18.19.20}\)
Cho A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
Chứng minh A<2
Tính nhanh:
a) A = 2\(^{2010}\) - 2\(^{2009}\) - 2\(^{2008}\) - 2\(^{2007}\) - ... - 2 - 1
b) B = 20 . 8 - 33 . 64 + 17 . 8 + 9 . 16 . 8 - 11 . 128
c) C = ( \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + ... + \(\dfrac{1}{2009.2010.2011}\) ) . \(\dfrac{2010.2011}{1010.527}\)
(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+....\(\dfrac{1}{8.9.10}\)) . x = \(\dfrac{22}{45}\)
Tìm x nha mọi người mình đang cần gấp lắm ạ
Chứng tỏ:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}=\dfrac{1}{4}\)
(Lưu ý dấu "." trong phân số là dấu nhân)
Giúp mình với các bạn!!!
