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![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(2^{x+3}+2^x=144\)
\(\Leftrightarrow2^x\cdot9=144\)
\(\Leftrightarrow2^x=16\)
hay x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2^0+2^1+2^2+2^3+2^4+2^5+\dots+2^{100}\\=(2^1+2^2)+(2^3+2^4)+(2^5+2^6)+\dots+(2^{99}+2^{100})+2^0\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+\dots+2^{99}\cdot(1+2)+1\\=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{99}\cdot3+1\\=3\cdot(2+2^3+2^5+\dots+2^{99})+1\)
Vì \(3\cdot(2+2^3+2^5+\dots+2^{99})\vdots3\)
\(\Rightarrow 3\cdot(2+2^3+2^5+\dots+2^{99})+1\) chia \(3\) dư 1
hay số dư của phép chia \(A\) cho \(3\) là \(1\).
A=2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100
A=1 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100
A=1 + (2^1 + 2^2) + (2^3 + 2^4) + ....+(2^99 + 2^100)
A=1 + 2.(1+2) + 2^3.(1+2)+....+2^99.(1+2)
A=1 + 2 . 3 + 2^3 . 3 +....+2^99 . 3
A=1 +3 .(2+2^3+..+2^99)
=> A:3 dư 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ko ghi đề
\(2A=2+2^2+...+2^{101}\\ 2A-A=2^{101}-1\\ =>A=2^{101}-1\)
Mấy cái khác cg lm như v (b thì 3b)
Nhớ đúng mk nhá
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a,\(A=3+3^2+3^3+...+3^{2010}\)
\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{2007}+3^{2008}+3^{2009}+3^{2010}\right)\)
\(=3\left(1+3+3^2+3^3\right)+....+3^{2007}\left(1+3+3^2+3^3\right)\)
\(=3.40+...+3^{2007}.40\)
\(=40\left(3+3^5+...+3^{2007}\right)⋮40\)
Vì A chia hết cho 40 nên chữ số tận cùng của A là 0
b,\(A=3+3^2+3^3+...+3^{2010}\)
\(3A=3^2+3^3+...+3^{2011}\)
\(3A-A=\left(3^2+3^3+...+3^{2011}\right)-\left(3+3^2+3^3+...+3^{2010}\right)\)
\(2A=3^{2011}-3\)
\(2A+3=3^{2011}\)
Vậy 2A+3 là 1 lũy thừa của 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 100 - 7 ( x - 5 ) = 31 + 33
100 - 7 ( x - 5 ) = 31 + 27
100 - 7 ( x - 5 ) = 58
7 ( x - 5 ) = 100 - 58
7 ( x - 5 ) = 42
x - 5 = 42 : 7
x - 5 = 6
=> x = 6 +5
=> x = 11
Vậy x = 11
b, 12 ( x - 1 ) : 3 = 43 + 23
12 ( x - 1 ) : 3 = 64 + 8
12 ( x - 1 ) : 3 = 72
12 ( x - 1 ) = 72 . 3
12 ( x - 1 ) = 216
x - 1 = 216 : 12
x - 1 = 18
=> x = 18 + 1
=> x = 19
Vậy x = 19
c, 24 + 5x = 75 : 73
24 + 5x = 72
24 + 5x = 49
5x = 49 - 24
5x = 25
=> x = 25 : 5
=> x = 5
Vậy x = 5
d, 5x - 206 = 24 . 4
5x - 206 = 16 . 4
5x - 206 = 64
5x = 64 + 206
5x = 270
=> x = 270 : 5
=> x = 54
Vậy x = 54
e, 125 = x3
53 = x3
=> x = 5
Vậy x = 5
g, 64 = x2
82 = x2
=> x = 8
Vậy x = 8
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
a) \(x=x^5\)
\(x^5-x=0\)
\(x\left(x^4-1\right)=0\)
\(x=0\) hoặc \(x^4-1=0\)
* \(x^4-1=0\)
\(x^4=1\)
\(x=1\)
Vậy x = 0; x = 1
b) \(x^4=x^2\)
\(x^4-x^2=0\)
\(x^2\left(x^2-1\right)=0\)
\(x^2=0\) hoặc \(x^2-1=0\)
*) \(x^2=0\)
\(x=0\)
*) \(x^2-1=0\)
\(x^2=1\)
\(x=1\)
Vậy \(x=0\); \(x=1\)
c) \(\left(x-1\right)^3=x-1\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)
*) \(x-1=0\)
\(x=1\)
*) \(\left(x-1\right)^2-1=0\)
\(\left(x-1\right)^2=1\)
\(x-1=1\) hoặc \(x-1=-1\)
**) \(x-1=1\)
\(x=2\)
**) \(x-1=-1\)
\(x=0\)
Vậy \(x=0\); \(x=1\); \(x=2\)