Ta có: \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{10^2}{10.11}\)
\(=\dfrac{2.2.3.3...10.10}{2.2.3.3.4...10.11}\)
\(=\dfrac{1}{11}\)
Vậy tích trên có giá trị \(=11.\)
A=\(\dfrac{1}{1.2^2}+\dfrac{1}{2.3^2}+\dfrac{1}{3.4^2}+...+\dfrac{1}{49.50^2}\)
B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
CM A<B
1. Chứng minh rằng:
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100}< 1\)
2. Chứng minh rằng:
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100-1}{100!}< 2\)
tính giá trị các biểu thức
a, A = \(\left(\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}\right):\left(\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}\right)\)
b, B = \(\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+....+\dfrac{181}{9.10}\)
CÁC BẠN ƠI GIÚP MÌNH VS????
HELP ME!!!!!!!!
Chứng minh rằng :\(\dfrac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98.99}=\dfrac{1}{2}\)
CMR:
\(B=\dfrac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98+99}=\dfrac{1}{2}\)
tính tỉ số \(\dfrac{A}{B}\) biết A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{101.102}\) và B=\(\dfrac{1}{52.102}\)+\(\dfrac{1}{53.101}\)+...+\(\dfrac{1}{102.52}\)+\(\dfrac{2}{77.154}\)
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...-\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\)
Ghi cách giải lun nha
Tính tổng
S = \(\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+....+\dfrac{2n+1}{[n\left(n+1\right)]^2}\)
Tính tổng S = \(\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)
