a/ \(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
b/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
c/ \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
d/ \(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}>4\)
Câu 1 :So sánh A và B
\(A=\dfrac{2^{2015} - 2}{2^{2016} + 1} B=\dfrac{2^{2016} - 2}{2^{2017} + 1}\)
Câu 2: Thực hiện phép tính
D = \(\dfrac{-1}{2} . 17,5 - \dfrac{2015}{2016}. 2018 + \dfrac{1}{2}.7,5+ \dfrac{2015}{2016}.2\)
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=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}\)
a)chứng minh rằng :\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)........+\(\dfrac{1}{100^2}< \dfrac{1}{2}\)
b)tính nhanh tổng S với S= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+......+\dfrac{1}{61.63}\)
các cao nhân gải giúp với ạ !!! iem đang cần gấp
giúp mk, please :)
\(\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{2017+\dfrac{2016}{6}+\dfrac{2015}{7}+...+\dfrac{1}{2021}}\)
A. \(\dfrac{1}{2020}\)
B. \(\dfrac{1}{2021}\)
C. \(\dfrac{1}{2019}\)
D. \(\dfrac{1}{2022}\)
chọn ra 3 ngừi nhanh nhứt:>>
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, \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}...+\dfrac{1}{9.10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{2}{5}\)
S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9}< 1-\dfrac{1}{9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{8}{9}\)
Vậy \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
a ) so sánh c và d biết :
C = \(\dfrac{1957}{2007}\) với D = \(\dfrac{1935}{1985}\)
b )hãy so sánh A và B
cho A = \(\dfrac{2016^{2016}+2}{2016^{2016}-1}\) và B = \(\dfrac{2016^{2016}}{2016^{2016}-3}\)
c ) so sánh M và N biết :
M = \(\dfrac{10^{2018}+1}{10^{2019}+1}\) ; N = \(\dfrac{10^{2019}+1}{10^{2020}+1}\)
tìm x e Q
a) \(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
b) \(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
c) \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
