\(\frac{1^2}{1^2-100+5000}+\frac{2^2}{2^2-200+5000}+........+\frac{99^2}{99^2-9900+5000}\)
1. tính:
A= \(\frac{1^2}{1^2-100+5000}+\frac{2^2}{2^2-200+5000}+...+\frac{99^2}{99^2-9900+5000}\)
giải nhanh nhé
Tính A= (1^2/1^2-100+5000)+(2^2/2^2-200+5000)+...+(99^2/99^2-9900+5000)
Tính A, biết:
A = \(\frac{1^2}{1^2-100+5000}\)+ \(\frac{2^2}{2^2-200+5000}\)+ \(\frac{3^2}{3^2-300+5000}\)+ ....+ \(\frac{99^2}{99^2-9900+5000}\)
Tính A = 12/(12-100+5000) + 22/(22-100+5000) + ... +992/(992-100+5000)
Ai hoc gioi giai giup minh
tính
M=(1-\(\frac{1}{1-\frac{2010}{1}}\)) . (2-\(\frac{1}{1-\frac{2010}{2}}\)) ...... (5000-\(\frac{1}{1-\frac{2010}{5000}}\))
Tính tổng
M = \(\left(1-\frac{1}{1-2010}\right)\left(2-\frac{1}{1-\frac{2010}{2}}\right)\left(3-\frac{1}{1-\frac{2010}{3}}\right)....\left(5000-\frac{1}{1-\frac{5000}{3}}\right)\)
Cho A = \(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}+\frac{4^2-3^2}{12^2}+.....+\frac{100^2-99^2}{9900^2}\)
Chứng minh A < 1
Tính tổng A = [ \(\frac{2017}{2}+\frac{2017}{6}+\frac{2017}{12}+...+\frac{2017}{9900}\)] : \(\frac{99}{100}\)
