Vì mỗi thừa số ( phân số ) trong tích D đều > 1
=> tích D < 1 < 10
=> D < 10
Vì mỗi thừa số ( phân số ) trong tích D đều > 1
=> tích D < 1 < 10
=> D < 10
Chứng minh \(\dfrac{1}{5}\)< \(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)+......+\(\dfrac{1}{99^2}\)+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{3}\)
Chứng minh rằng :
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\)
Chứng tỏ rằng :\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)=2
Chứng tỏ rằng:\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\)
Trình bày ra dùm mình nha!!Giúp em nha!!!tính D =\(\dfrac{100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....+\dfrac{99}{100}}\)
Cho A=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
Chứng minh rằng 0,2<0,4.
chứng tỏ rằng
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}>\dfrac{99}{202}\)
Bài 1: Tìm số nguyên x, biết:
a)\(\dfrac{6}{x-3}\) = \(\dfrac{2}{3}\)
b) \(\dfrac{14}{13}\) = \(\dfrac{-28}{10-x}\)
c) \(\dfrac{1}{5}\) = \(\dfrac{x:4-1}{10}\)
d) \(\dfrac{x}{4}\)= \(\dfrac{1}{x}\)
e) \(\dfrac{x-2}{50}\) = \(\dfrac{2}{x-2}\)
giúp ưm
Cho \(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}\). Chứng minh \(B< 1\dfrac{3}{4}\)