Cho H=\(2^{2018}-2^{2017}-....-2-1\)
Tính \(2018^H\)
K
Khách
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NP
2
30 tháng 12 2022
Đặt G=2^2017+2^2016+...+2+1
=>2G=2^2018+2^2017+...+2^2+2
=>G=2^2018-1
=>H=2^2018-2^2018+1=1
=>2018^H=2018
GH
1
21 tháng 4 2018
H=22018- 22017- 22016 - ... - 2 - 1
⇒2H=22019 - 22018 - 22017 - ... - 4 -2
⇒2H - H = 22019 -1
⇒ 2018H = 2^(2^2019-1)
TT
1
XO
18 tháng 2 2020
\(A=\frac{1}{2018}+\frac{2}{2017}+...+\frac{2017}{2}+2018\)
\(=\left(\frac{1}{2018}+1\right)+\left(1+\frac{2}{2017}\right)+...+\left(\frac{2017}{2}+1\right)+1\)(2018 số hạng 1)
\(=\frac{2019}{2018}+\frac{2019}{2017}+...+\frac{2019}{2}+\frac{2019}{2019}=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)\)
Mà \(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
=> Khi đó : \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
21 tháng 2 2021
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27 tháng 12 2017
A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+..........+\frac{2018}{2017^2+2017}\)
>\(\frac{2018}{2017^2+2017}+\frac{2018}{2017^2+2017}+........+\frac{2018}{2017^2+2017}\)
\(=\frac{2018}{2017^2+2017}.2017=\frac{2018.2017}{2017\left(2017+1\right)}=1\) (1)
Lại có:A<\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+1}+.........+\frac{2018}{2017^2+1}\)
\(=\frac{2018}{2017^2+1}.2017=\frac{2018.2017}{2017^2+1}=\frac{2017.\left(2017+1\right)}{2017^2+1}\)
\(=\frac{2017^2+2017}{2017^2+1}=\frac{2017^2+1+2016}{2017^2+1}=1+\frac{2016}{2017^2+1}< 2\) (2)
Từ (1) và (2) suy ra:1 < A < 2
Vậy A không phải là số nguyên
14 tháng 10 2021
Đặt \(2017=a\)
\(A=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2a+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2\left(a+1\right)\cdot\dfrac{a}{a+1}+\left(\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1-\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\left|a+1-\dfrac{a}{a+1}\right|+\dfrac{a}{a+1}\)
Ta có \(\dfrac{a}{a+1}< 1\Leftrightarrow a+1-\dfrac{a}{a+1}>0\)
\(\Leftrightarrow A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2018\)
A
25 tháng 8 2021
\( S =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1} {2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right) \)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\) \(\Rightarrow S=P\)\)
25 tháng 8 2021
\(B=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(B=1+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{1}{2018}+1\right)\)
\(B=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(B=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
ta có \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
\(H=2^{2018}-\left(2^{2017}+2^{2016}+...+1\right)\)
\(H=2^{2018}-\left(\dfrac{2^{2018}+2^{2017}+...+2}{2}\right)\)
\(H=2^{2018}-\left(2^{2018}+2^{2017}+...+2-2^{2017}-2^{2016}-...-1\right)\)
\(H=2^{2018}-2^{2018}+1=1\)
Vậy \(2018^H=1\)