Chứng tỏ rằng:
\(\dfrac{\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{199.200}}{\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}}=1\)
Giúp mình với📖
Bài 1 : Tính tổng sau
a) \(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
b) \(B=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+....+\dfrac{1}{23.24}+\dfrac{1}{24.25}\)
c) \(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
1.Tính
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)
B=\(\dfrac{3}{5.6}+\dfrac{3}{6.7}+\dfrac{3}{7.8}+.....+\dfrac{3}{101.102}\)
C=\(\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}\)
D=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}\)
Bài 1: Thực hiện các phép tính:
d) 3,15+2,4=5,55
e) \(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}\)
f) 1,25.3,6+3,6.8,75
h) B= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
so sánh A và B biết:
A=\(\dfrac{2^{2018}}{2^{2018}+3^{2019}}\)+\(\dfrac{3^{2019}}{3^{2019}+5^{2020}}\)+\(\dfrac{5^{2020}}{5^{2020}+2^{2018}}\)
B=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{2019.2020}\).
Chứng tỏ rằng:
a, \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}< 1\)
b, \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
Tính tổng:
\(M=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
\(\dfrac{1}{1.2}+\dfrac{2}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.(x+1)}=\dfrac{2021}{2022}\)