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P = 50 + 51 + 52 + .... + 52016
5P = 5( 50 + 51 + 52 + .... + 52016 )
= 51 + 52 + 53 + .... + 52017
5P - P = ( 51 + 52 + 53 + .... + 52017 ) - ( 50 + 51 + 52 + .... + 52016 )
4P = 52017 - 1
=> P = ( 52017 - 1 ) : 4
=> Q - P = 52017 : 4 - ( 52017 - 1 ) : 4
= [ 52017 - ( 52017 - 1) ] : 4
= ( 52017 - 52047 + 1 ) : 4
= 1/4
\(C=5^{2018}+\frac{1}{5^{2017}+1}=\left(5^{2017}+1\right)+\frac{1}{5^{2017}+1}\)
\(D=5^{2018}+\frac{1}{5^{2018}+1}=\left(5^{2017}+1\right)+\left(1+\frac{1}{5^{2017}+2}\right)\)
Do \(\frac{1}{5^{2017}+1}< 1+\frac{1}{5^{2017}+2}\)
Nên \(C< D\)
Ta có : C = \(\frac{5^{2018}+1}{5^{2017}+1}\)
=> \(\frac{C}{5}=\frac{5^{2018}+1}{5^{2018}+5}=1-\frac{4}{5^{2018}+5}\)
Lại có D = \(\frac{5^{2019}+1}{5^{2018}+1}\)
=> \(\frac{D}{5}=\frac{5^{2019}+1}{5^{2019}+5}=1-\frac{4}{5^{2019}+5}\)
Vì \(\frac{4}{5^{2018}+5}>\frac{4}{5^{2019}+5}\Rightarrow1-\frac{4}{5^{2018}+5}< 1-\frac{4}{5^{2019}+5}\Rightarrow\frac{C}{5}< \frac{D}{5}\Rightarrow C< D\)
Ta có:
A = 1 + 5 + 52 + 53 + 54 + ...+ 52017
A = \(\frac{5^{2017}-1}{5-1}\)
B = \(\frac{5^{2018}-1}{2-1}\)
=> \(4A=\frac{5^{2017}-1}{4}.4=5^{2017}-1< B=5^{2018}-1\)
Vậy 4A < B
Ta có: 5A=5(1+5+52+....+52017)
5A=5+52+53+....+52018
5A-A=(5+52+53+...+52018)-(1+5+52+....+52017)
4A=52018-1
Vì 4A=52018-1. Mà 52018-1=52018-1
Suy ra:4A=B
\(\left(x-1\right)^{10}=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
E=-5+5^2-5^3+5^4-........-5^2017+5^2018
5E=-5^2+5^3-5^4+.........+5^2018-5^2019
5E+E=-5^2019-5
E=(-5^2019-5):4
vậy E=(-5^2019-5):4
5P=5+5^1+5^2+...+5^2017+5^2018
5P-1P=(5+5^2+....+5^2017+5^2018)-(5^0+5*1+...+5^2017)
4P=5^2018-1
P=(5^2018-1):4
=> Q-P=1/4
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