(3x - 1)3 = 125
(3x - 1)3 = 53
=>3x - 1 = 5
3x = 5 + 1
3x = 6
x = 6 : 3
x = 2
A = 1+5+52+53+...+597+598
A = (1 + 5 + 52) + (53 + 54 + 55) + ... + (596 + 597 + 598)
A = 1(1 + 5 + 52) + 53(1 + 5 + 52) + ... + 596(1 + 5 + 52)
A = 1.31 + 53.31 + ... + 596.31
A = 31(1 + 53 + ... + 596)
Vì 31(1 + 53 + ... + 596) \(⋮\)nên A \(⋮\)31
Vậy A \(⋮\)31
a, \(\left(3x-1\right)^3=125\Leftrightarrow\left(3x-1\right)^3=5^3\)
\(\Rightarrow3x-1=5\Rightarrow3x=5+1\Rightarrow3x=6\Rightarrow x=6\div3=2\)
Vậy x = 2
b, Xét dãy số mũ : 0;1;2;3;...;97;98
Số số hạng của dãy số trên là :
\(\left(98-0\right)\div1+1=99\) ( số )
Ta được số nhóm là :
\(99\div3=33\) ( nhóm )
Ta có : \(A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{96}+5^{97}+5^{98}\right)\) (33 nhóm )
\(A=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+...+5^{96}\left(1+5+5^2\right)\)
\(A=1.31+5^3.31+...+5^{96}.31=\left(1+5^3+...+5^{96}\right).31\)
Mà : \(31⋮31;1+5^3+...+5^{96}\in N\Rightarrow A⋮31\) (đpcm)
\(\left(3x-1\right)^3=125\)
\(\Rightarrow3x-1=5\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Ta có: \(A=1+5+5^2+...+5^{97}+5^{98}\)
\(\Rightarrow A=\left(1+5+5^2\right)+...+\left(5^{96}+5^{97}+5^{98}\right)\)
\(\Rightarrow A=31+...+5^{96}\left(1+5+5^2\right)\)
\(\Rightarrow A=31+...+5^{96}.31\)
\(\Rightarrow A=\left(1+...+5^{96}\right).31⋮31\)
Vậy\(A⋮31\)
\(a,\left(3x-1\right)^3=125\)
\(\left(3x-1\right)^3=5^3\)
\(\left(3x-1\right)=5\)
\(3x\) \(=5+1\)
\(3x\) \(=6\)
\(x=6:3\)
\(x=2\)
\(A=1+5+5^2+5^3+....+5^{97}+5^{98}\)
\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+....+\left(5^{96}+5^{97}+5^{98}\right)\)
\(=31+\left(5^3+5^4+5^5\right)+....+\left(5^{96}+5^{97}+5^{98}\right)\)
\(=31+5^3\left(1+5+5^2\right)+....+5^{96}\left(1+5+5^2\right)\)
\(=31+5^3.31+....+5^{96}.31\)
\(=31.\left(1+5^3+....+5^{96}\right)\) chia hết cho 31
