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

b)   2^x+3 +2^x =144

     2^X . 2^3 + 2^x =144

      2^x.( 2^3+1) =144

      2^x. 9            =144

       2^x                =144:9

      2^x                = 16

=> 2^x                 = 2^4

-Vậy  x = 4

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 \(x=0\); \(x=1\); \(x=2\)

Bài 3 :

a) 4.(x-5) - 2 ³=2⁴.3

4x-20-8=48

4x=76

x=19

b) 4.x³+15=271

4.x³=256

x³=64

=> x=4

c) ( 2x-3)²= 169

=> 2x-3= 13

2x=16

x=8

 Chúc bạn học tốt !

4*(x-5) - 2^3 = 2^4*3 

4*(x-5) - 8 = 16*3 

4*(x-5) - 8 = 48 

4*(x-5) = 48 + 8 

4*(x-5) = 56 

x- 5 = 56 : 4 

x - 5 = 14 

x = 14 + 5 

x = 19 

4*x^3 + 15 = 271 

4*x^3 = 271 - 15 

4*x^3 = 256 

x^3 = 256 : 4 

x^3 = 64 

x^3 = 4^3 

x = 4 

Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3

Bài 1 :

a) 2^x-64=2⁶

=> 2^x-2⁶=2⁶

2^x=2⁶+2⁶ (2⁶⁺⁶)

2^x = 2¹²

<=> x = 12

b) 2^x:16=128

=> 2^x.2⁴ = 2⁷

2^x = 2⁷:2⁴ (2^7-4)

2^x = 2³

<=> x=3

Bài 2 :

b) A= 8² = (2³)² = 2⁶  

B= 2⁶

=> A = B ( 2⁶=2⁶ )

c) A= 3⁵ = 243

B = 2⁸ = 256

=> A < B ( 243 < 256 )

a/

\(x^3-4x^2-\left(x-4\right)=0\)

\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)

b/

\(x^5-9x=0\)

\(\Leftrightarrow x\left(x^4-9\right)=x\left(x^2-3\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

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

c/

\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)

\(\Leftrightarrow x^4\left(x-1\right)^2-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^4-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^2-2\right)\left(x^2+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{2}\end{matrix}\right.\)

Sos

giúp với mn

a) 2^x : 4 = 128

=> 2x  = 512

=> 2^x  = 2⁹

=> x = 9

b) x¹⁵ = x

=> x¹⁵  = x¹⁵

=> x = 1 hoặc 0

c) ( 2x + 1 )³ = 125

=> ( 2x + 1 )³ = 5³

=> 2x + 1 = 5

=> 2x = 4

=> x = 2

a) 2^x : 4 = 128

=> 2^x = 128 . 4

=> 2^x = 512

=> 2^x = 2⁹

=> x = 9

b) x¹⁵ = x

=> x¹⁵ = x¹⁵

=> x = 1 hoặc 0

c) ( 2x + 1 )³ = 125

=> ( 2x + 1 )³ = 5³

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2