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Ta có: 5x+x+x+3 ≤ 1018 : 218
⇒53x+3 ≤ 518
⇒3x + 3 ≤ 18 ⇒ 3x+3 ≤ 15
⇒ x ≤ 5
Vậy x ϵ (0;1;2;3;4;5)
![](https://rs.olm.vn/images/avt/0.png?1311)
5x . 5x+1 . 5x+2 = 1000...0 ( có 18 chữ số 0 ) : 218
5x + ( x + 1 ) + ( x + 2 ) = 1018 : 218
5x + x + 1 + x + 2 = ( 10 : 2 )18
5 x + x + x + 1 + 2 = 518
5 3x + 3 = 518
=> 3x + 3 = 18
3x = 18 - 3
3x = 15
x = 15 : 3
x = 5.
Vậy x = 5.
Theo bài ra, ta có:
5x . 5x+1 . 5x+2 = 1018 : 218
\(\Rightarrow\) 5x . 5x . 5 . 5x . 52 = (10:2)18
\(\Rightarrow\) (5x . 5x . 5x).(5.52) = 518
\(\Rightarrow\) 53x . 53 = 518
\(\Rightarrow\) 53x = 518 : 53
\(\Rightarrow\) 53x = 515
\(\Rightarrow\) 3x = 15
\(\Rightarrow\) x = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2x . 2x + 1 . 2x + 2 = 100...00 : 518 (18 chữ số 0)
=> 23x + 3 = 1018 : 518
=> 23x + 3 = (10 : 5)18
=> 23x + 3 = 218
=> 3x + 3 = 18
=> x = 5
2x . 2x+1 . 2x+2 = 100...0 : 518(18 chữ số 0)
2x+x+1+x+2 = 1018 : 518
23x+3 = ( 10 : 5 )18
23x+3 = 218
-> 3x+3 = 18
3x = 18 - 3 = 15
x = 15 : 3 = 5
Vậy x = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2^x.2^{x+1}.2^{x+2}=10^{18}:5^{18}\)
\(\Rightarrow2^{x+x+1+x+2}=2^{18}\Rightarrow x+x+1+x+2=18\Rightarrow3x+3=18\Rightarrow3x=15\Rightarrow x=5\)
2x.2x + 1.2x + 2 \(\le\) 1000...00 : 259
<=> 2x + x + 1 + x + 2 \(\le\) 1009 : 259
<=> 23x + 3 \(\le\) 49
<=> 23x + 3 \(\le\) 218
<=> 3x + 3 \(\le\) 18
<=> x \(\le\) 5
Mà x \(\in\) N => x \(\in\) {0; 1; 2; 3; 4; 5}
thank you