bài này dễ ợt

a,\(4^n.2^n=512\)

\(\Rightarrow2^{2n}.2^n=512\Rightarrow2^{3n}=2^9\Rightarrow3n=9\Rightarrow n=3\)

b,\(3^n+3^{n+3}=252\)( sửa đề ) 

\(\Rightarrow3^n.\left(1+3^3\right)=252\Rightarrow3^n.28=252\Rightarrow3^n=9\Rightarrow n=2\)

c,\(2.3^{2x+2}=18\)

\(\Rightarrow3^{2n+2}=9\Rightarrow2n+2=2\Rightarrow n=0\)

d,\(x^2=2^3+3^2+4^3\)

\(\Rightarrow x^2=8+9+64\Rightarrow x^2=81\Rightarrow x^2=9^2=\left(-9\right)^2\Rightarrow x=9\)hoặc \(x=-9\)

e,\(x^5=x^9\)

\(\Rightarrow x^9-x^5=0\Rightarrow x^5.\left(x^4-1\right)=0\Rightarrow\hept{\begin{cases}x^5=0\\x^4-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\\x=-1\end{cases}}}\)

f,\(\left(x-4\right)^3=\left(x-4\right)^{10}\)

\(\Rightarrow\left(x-4\right)^{10}-\left(x-4\right)^3=0\Rightarrow\left(x-3\right)^3.\left[\left(x-3\right)^7-1\right]=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-3\right)^3=0\\\left(x-3\right)^7=1\end{cases}\Rightarrow\hept{\begin{cases}x-3=0\\x-3=1\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\x=4\end{cases}}}\)

a) 2^x . 4 = 128

2^x = 128 : 4

2^x = 32

2^x = 2⁵

=> x = 5

b) 2^x . 2⁴ = 2⁶

=> x + 4 = 6

x = 6 - 4

x = 2

c) 5x + x = 39 - 3¹¹ : 3⁹

6x = 39 - 3²

6x = 39 - 9

6x = 30

x = 30 : 6

x = 5

d) 9^{x - 1} = 81

9^{x - 1} = 9²

=> x - 1 = 2

x = 2 + 1

x = 3

e) 6x = 5²¹ : 5¹⁹ + 3 . 2² - 7⁰

6x = 5² + 3 . 4 - 1

6x = 25 + 12 - 1

6x = 37 - 1

6x = 36

x = 36 : 6

x = 6

x(3-y)+3y=8=> x(3-y)+3y-9=8-9=>x(3-y)-3(3-y)=-1=>(x-3)(3-y)=-1

Ta co bang sau

x-3   1   -1

3-y  -1    1

x      4    2

y      4    2

Câu 4
Đặt \(A=3+3^2+...+3^{20}\)

\(\Rightarrow A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{19}+3^{20}\right)\)

\(\Rightarrow A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{19}\left(1+3\right)\)

\(\Rightarrow A=3.4+3^3.4+...+3^{19}.4\)

\(\Rightarrow A=\left(3+3^3+...+3^{19}\right).4⋮4\)

\(\Rightarrow A⋮4\left(đpcm\right)\)

\(A=3+3^2+...+3^{20}\)

\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\)

\(\Rightarrow A=3\left(1+3+3^2+3^3\right)+...+3^{17}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=3.40+...+3^{17}.40\)

\(\Rightarrow A=\left(3+...+3^{17}\right).40⋮40\)

\(\Rightarrow A⋮40\left(đpcm\right)\)

Câu 3:

Giải:
a) \(5⋮x-5\)

\(\Rightarrow x-5\in\left\{1;5\right\}\)

+) \(x-5=1\Rightarrow x=6\)

+) \(x-5=5\Rightarrow x=10\)

Vậy \(x\in\left\{6;10\right\}\)

b) Ta có: \(x+3⋮x-3\)

\(\Rightarrow\left(x-3\right)+6⋮x-3\)

\(\Rightarrow6⋮x-3\)

\(\Rightarrow x-3\in\left\{1;2;3;6\right\}\)

\(\Rightarrow x\in\left\{4;5;6;9\right\}\)

Vậy \(x\in\left\{4;5;6;9\right\}\)

\(A=3^1+3^2+...+3^{2006}\)

\(3A=3^2+3^3+...+3^{2007}\)

\(3A-A=\left(3^2+3^3+...+3^{2007}\right)-\left(3^1+3^2+...+3^{2006}\right)\)

\(2A=3^{2007}-3\)

\(A=\frac{3^{2007}-3}{2}\)

\(2A+3=3^x\)

\(\left(3^{2007}-3\right)+3=3^x\)

\(3^{2007}+\left(-3\right)+3=3^x\)

\(3^{2007}+\left[\left(-3\right)+3\right]=3^x\)

\(\Rightarrow3^{2007}=3^x\)

\(\Rightarrow x=2007\)

a) A bằng 3¹+3²+3³+3⁴+...+3²⁰⁰⁶

3A bằng 3.(3¹+3²+3³+3⁴+...+3²⁰⁰⁶)

3A bằng 3²+3³+3⁴+3⁵+...+3²⁰⁰⁷

3A-A bằng (3²+3³+3⁴+3⁵+...+3²⁰⁰⁷) - (3¹+3²+3³+3⁴+...+3²⁰⁰⁶)

  2A   bằng        3²⁰⁰⁷-3¹

    A   bằng    (3²⁰⁰⁷-3) : 2 

b) 2A+3 bằng 3^x

Thay 2A bằng 3²⁰⁰⁷-3, ta có :

2A+3 bằng 3^x

3²⁰⁰⁷-3+3 bằng 3^x

3²⁰⁰⁷ bằng 3^x

suy ra x bằng 2007

Vậy x bằng 2007

\(\left(3^2\right)^2+2^x=5\left(5+2^2.3\right)\)

\(\Rightarrow81+2^x=5.17\)

\(\Rightarrow81+2^x=85\)

\(\Rightarrow2^x=85-81=4=2^2\Rightarrow x=2\)

\(5^{x-1}+5=150\Rightarrow5^{x-1}=150-5=145\)

Mà không có lũy thừa nào có cơ số là 5 mà kết quả bằng 145 nên x không thỏa mãn điều kiện trên.

\(\left(3-x\right)^6=\left(3-x\right)^4\)

\(\Rightarrow3-x=\left\{-1;0;1\right\}\)

\(\Rightarrow x=\left\{4;3;2\right\}\)

9^2+2^x=5(5+4.3)

81+2^x=5(5+12)

81+2^x=5.17

81+2^x=85

2^x=85-81

2^x=4

vi 4= 2^2 . x=2

mk ko hiểu đề