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\(A=\frac{2017^{2018+1}}{2017^{2018-3}}\)và \(B=\frac{2017^{2018-1}}{2017^{2018-5}}\)
Có \(A=\frac{2017^{2019}}{2017^{2015}}\)và \(B=\frac{2017^{2017}}{2017^{2013}}\)
Mà\(\frac{2017^{2019}}{2017^{2015}}>\frac{2017^{2018}}{2017^{2015}}\)và\(\frac{2017^{2017}}{2017^{2013}}>\frac{2017^{2017}}{2017^{2015}}\)
Vì \(\frac{2017^{2018}}{2017^{2015}}>\frac{2017^{2017}}{2017^{2015}}\)
Vậy A>B
(5^2018+5^2018+5^2018+5^2018) + 5^2018 -5x=0
5^2018+5^2018+5^2018+5^2018+5^2018-5x =0
5(5^2018)-5x =0
5x =5(5^2018)-0
5x =5(5^2018)
Suy ra x= 5^2018
Vậy: x= 5^2018
Sai đề câu E sửa lại 95 hoặc 93 vì đây là dãy số mũ lẻ. Ta có :
\(E=3+3^3+3^5+3^7+...+3^{95}\)
\(\Rightarrow\) \(9E=3^3+3^5+3^7+3^9+...+3^{95}+3^{97}\)
\(\Rightarrow\) \(8E=3^{97}-3\)
\(\Rightarrow\) \(E=\frac{3^{97}-3}{8}\)
\(E=3+3^3+3^5+3^7+.......+3^{95}\)
\(\Rightarrow9E=3^3+3^5+3^7+3^9+...+3^{97}\)
\(\Rightarrow9E-E=\left(3^3+3^5+3^7+3^9+....+3^{97}\right)-\left(3+3^3+3^5+3^7+.....+3^{95}\right)\)
\(\Rightarrow8E=3^{97}-3\)
\(\Rightarrow E=\frac{3^{97}-3}{8}\)
\(F=1+2018+2018^2+......+2018^{2017}\)
\(=2018^0+2018^1+2018^2+....+2018^{2017}\)
\(\Rightarrow2018F=2018^1+2018^2+2018^3+....+2018^{2018}\)
\(\Rightarrow2018F-F=\left(2018^1+2018^2+2018^3+....+2018^{2018}\right)-\left(2018^0+2018^1+2018^2+....+2018^{2017}\right)\)
\(\Rightarrow2017F=2018^{2018}-1\)
\(\Rightarrow F=\frac{2018^{2018}-1}{2017}\)
\(A=5+3^2+3^3+...+3^{2018}\)
\(3A=15+3^3+3^4+...+3^{2019}\)
\(3A-A=\left(15+3^3+3^4+...+3^{2019}\right)-\left(5+3^2+3^3+...+3^{2018}\right)\)
\(2A=1+3^{2019}\)
\(2A-1=3^{2019}\)
Suy ra \(n=2019\).
\(5^6:5^4+2^3+2^2-1^{2018}\)
\(=5^2+8+4-1\)
\(=25+11\)
\(=36\)
56 : 54 + 23 + 22 - 12018
= 52 + 23 + 22 - 12018
= 25 + 8 + 4 - 1
= 36
Học tốt nhé bạn !