\(A=5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9\)

\(A=5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9\)

\(A=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}\)

\(A=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}\)

\(A=2^{29}\cdot3^{18}\cdot\left(5\cdot2^1\cdot1-1\cdot3^2\right)\)

\(A=2^{29}\cdot3^{18}\cdot\left(5-9\right)\)

\(A=-2^2\cdot2^{29}\cdot3^{18}\)

\(A=-2^{31}\cdot3^{18}\)

_______________

\(B=5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6\)

\(B=5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6\)

\(B=5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)

\(B=2^{28}\cdot3^{18}\cdot\left(5\cdot1\cdot3-7\cdot2\cdot1\right)\)

\(B=2^{28}\cdot3^{18}\cdot\left(15-14\right)\)

\(B=2^{28}\cdot3^{18}\)

Ta có: \(A:B\)

\(=\left(-2^{31}\cdot3^{18}\right):\left(2^{28}\cdot3^{18}\right)\)

\(=\left(-2^{31}:2^{28}\right)\cdot\left(3^{18}:3^{18}\right)\)

\(=-2^3\cdot1\)

\(=-8\)

100%

b: Ta có: \(2^{x+3}+2^x=144\)

\(\Leftrightarrow2^x\cdot9=144\)

\(\Leftrightarrow2^x=16\)

hay x=4

a) (x ^ 54)^2 = x                                         

         x^108  = x

Để: x^108  = x 

=> x=0 hoặc x=1

a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)

b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)

c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)

d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)

f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)

g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)

h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)

i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1=1^{49}\end{matrix}\right.\)  \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

4ⁿ  =  4096 

4ⁿ = 2¹²

n = 12

5ⁿ = 15625 

5ⁿ = 5⁶

n   = 6

6ⁿ⁺³ = 216

6ⁿ⁺³ = 2³.3³

6ⁿ⁺³ = 6³

n + 3 = 3

a, x= 3

b, ko có x thỏa mãn

c, x= 3

d, x= 2

a: x=3

b: \(2x-1=2\)

hay \(x=\dfrac{3}{2}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

b: =>12-x=-7

hay x=19

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

a) \(3^2.x+2^3.x=51\)

\(\Leftrightarrow x\left(3^2+2^3\right)=51\)

\(\Leftrightarrow17x=51\)

\(\Leftrightarrow x=3\)

Vậy

b) \(6^2.2-\left(84-3^2.x\right):7=69\)

\(\Leftrightarrow\left(84-3^2.x\right):7=3\)

\(\Leftrightarrow84-3^2.x=21\)

\(\Leftrightarrow3^2.x=63\)

\(\Leftrightarrow x=7\)

Vậy

a) \(\left(x-1\right)^2=49\)

\(\Rightarrow\left(x-1\right)^2=7^2=\left(-7\right)^2\)

\(\Rightarrow x-1=7\)                        hoặc                           \(x-1=-7\)

\(x=7+1=8\)                                                                \(x=-7+1=-6\)

Vậy x = 8 hoặc x = - 6

b) \(3\cdot\left(13-x\right)^2=27\)

\(\left(13-x\right)^2=27\div3=9\)

\(\Rightarrow\left(13-x\right)^2=3^2=\left(-3\right)^2\)

\(\Rightarrow13-x=3\)                                          hoặc                                    \(13-x=-3\)

\(x=13-3=10\)                                                                                            \(x=13+3=16\)

Vậy x = 10 hoặc x = 16

c) \(164-\left(15-x\right)^3=100\)

\(\left(15-x\right)^3=164-100=64\)

\(\Rightarrow\left(15-x\right)^3=4^3\)

\(\Rightarrow15-x=4\)

\(x=15-4=11\)

Vậy x = 11

d) \(\left(x+3\right)^3-15=210\)

\(\left(x+3\right)^3=210+15=225\)

\(\Rightarrow\left(x+3\right)^3=...\)

Tương tự mũ lẻ cậu nhé

e) \(x^2\div4=16\)

\(x^2=16\cdot4=64\)

\(\Rightarrow x^2=8^2=\left(-8\right)^2\)
Vậy x = 8 hoặc x = - 8

a/\(\left(x-1\right)^2\)=49

\(\left(x-1\right)^2\)=\(7^2\)

=>x-1=7

x=7+1

x=8

b/3.\(\left(13-x\right)^2\)=27

\(\left(13-x\right)^2\)=27:3

\(\left(13-x\right)^2\)=9

\(\left(13-x\right)^2\)=\(3^2\)

=>13-x=3

x=13-3

x=10

c/164-\(\left(15-x\right)^3\)=100

\(\left(15-x\right)^3\)=164-100

\(\left(15-x\right)^3\)=64

\(\left(15-x\right)^3\)=\(4^3\)

=>15-x=4

x=15-4

x=11

d/\(\left(x+3\right)^3\)-15=210

\(\left(x+3\right)^3\)=210+15

\(\left(x+3\right)^3\)=225

sai đề bài câu d hay sao ý bạn ạ 

chỉ có \(\left(x+3\right)^2\)thì mới tính được

e/\(x^2\):4=16

\(x^2\)=16.4

\(x^2\)=64

\(x^2\)=\(8^2\)

=>x=8

a) 3(x - 1) - 1 = 26

=> 3x - 3 = 26 + 1

=> 3x - 3 = 27

=> 3x = 27 + 3

=> 3x = 30

=> x = 30 : 3

=> x = 10

b) |x + 4| - 9 = (-2)³

=> |x + 4| - 9 = -8

=> |x + 4| = -8 + 9

=> |x + 4| = 1

=> \(\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\) 

=> \(\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)

Vậy ...

c) 2x - 5 \(⋮\)x - 1

<=> 2(x - 1) - 3 \(⋮\)x - 1

<=> 3 \(⋮\)x - 1

<=> x - 1 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng :

Vậy ...