a) 2^x . 4 = 128

<=> 2^x = 32 

<=> 2^x = 2⁵

<=> x = 5

b) x¹⁵ = x¹

<=> x¹⁵ - x = 0

<=> x(x¹⁴ - 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

c) (2x + 1)³ = 125

<=> (2x + 1)³ = 5³

<=> 2x + 1 = 5

<=> 2x = 4

<=> x = 2

d) (x - 5)⁴ = (x - 5)⁶

<=> (x - 5)⁶ - (x - 5)⁴ = 0

<=> (x - 5)⁴[(x - 5)² - 1] = 0

<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

Khi (x - 5)⁴ = 0 => x - 5 = 0 => x = 5

Khi (x - 5)² - 1 = 0 <=> (x - 5)² = 1² <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

a, 2x . 4 = 128

=> 2x = 128 : 4 = 32

=> x = 32 : 2 = 16

Vậy x = 16

b, x15 = x 1 => Sai đề

c, (2x + 1)³ = 125

=> ( 2x + 1 ) = 5³

=> 2x + 1 = 5 

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2

2^x . 4 = 128

2^x      = 128 : 4

2^x      = 32

2^x      = 2 . 2 . 2 . 2 . 2

2^x        = 2⁵

x       = 5

(2x + 1)³ = 125

(2x + 1)³ = 5 . 5 . 5

(2x + 1)³ = 5³

2x + 1    = 5

2x         = 5 - 1 

2x         = 4

x          = 4 : 2

x          = 2

x¹⁵ = x

x = 1

(x - 5)⁴ = (x - 5)⁶

x = 6

a,x=5

b,x=2

c,x=6

tk mk nhé

125=5 mũ 3; suy ra 2x+1=5;2x=4 suy ra x=2

x=1 vì 1 mũ n luôn =1

chịu nhé

a) x=5

b) x=0 hoặc 1

c) x=2

d) x=6 hoặc 5

e) x=0 hoặc 1

f) x=8

a, 2^x . 4 = 128

2^x = 128 : 4

2^x = 32

2^x = 2⁵

=> x = 5

b, x¹⁵ = x¹

=> x¹⁵ - x = 0

x . ( x¹⁴ - 1 ) = 0

=> x = 0 hoặc x¹⁴ - 1 = 0 

=> x = 0 hoặc x = 1

c, (2x + 1)³ = 125

( 2x + 1 )³ = 5³

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2

d, (x – 5)⁴ = (x - 5)⁶

=> ( x - 5 )⁶ - ( x - 5 )⁴ = 0

=> ( x - 5 )⁴ . [ ( x - 5 )² - 1 ] = 0

=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

e, x¹⁰ = x

x¹⁰ - x = 0

x . ( x⁹ - 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

f, (2x -15)⁵ = (2x -15)³

( 2x - 15 )⁵ - ( 2x - 15 )³ = 0

( 2x - 15 )³ . [ ( 2x - 15 )² - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow}\orbr{\begin{cases}x\text{ không tồn tại}\\x=8\end{cases}}}\)

a) \(2^x.4=128\)

\(\Rightarrow2^x=128:4=32\)

Mà \(32=2^5\)

\(\Rightarrow2^x=2^5\)

Vậy x = 5

b) \(x^{15}=x\)

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

c) Ta có: \(125=5^3\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=5-1=4\)

\(\Rightarrow x=4:2=2\)

Vậy x = 2

d) Ta có: \(\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Rightarrow x=5\)

Ủng hộ tớ nha?

\(2^x.4=128\)

\(\Rightarrow2^x.2^2=2^7\)

\(\Rightarrow x+2=7\)

\(\Rightarrow x=5\)

\(x^{15}=x\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=+-1\end{cases}}\)

\(\left(2x+1\right)^3=125\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

a) 2x.4=128

⇒2x=128:4=32

Mà 32=25

⇒2x=25

Vậy x = 5

b) x15=x

⇒x={0;1;−1}

c) Ta có: 125=53

⇒2x+1=5

⇒2x=5−1=4

⇒x=4:2=2

Vậy x = 2

d) Ta có: (x−5)4=(x−5)6

⇒x=5

a ) 2^x . 4 = 128

2^x = 128 : 4

2^x = 32

2^x = 2⁵

=> x = 5

Vậy x = 5

b ) x¹⁵ = x

=> x¹⁵ : x = 1

x¹⁴ = 1

x¹⁴ = 1¹⁴

=> x = 1

Vậy x = 1

c ) ( 2x + 1 )³ = 125

( 2x + 1 )³ = 5³

=> 2x + 1 = 5

2x = 5 - 1

2x = 4

=> x = 4 : 2

x = 2

Vậy x = 2

d ) ( x - 5 )⁴ = ( x - 5 )⁶

( x - 5 )⁶ - ( x - 5 )⁴ = 0

( x - 5 )⁴ . [ ( x - 5 )² - 1 ] = 0

=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\text{[}\left(x-5\right)^2-1=0\end{cases}}\)=> \(\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)=> \(\orbr{\begin{cases}x=5\\x-5=1\end{cases}}\)=> \(\orbr{\begin{cases}x=5\\x=6\end{cases}}\)

Vậy x thuộc { 5 ; 6 }

a/ 2^x .4 =128

    2^x =128 : 4

    2^x = 32

     mak 2^4 =32      => x= 5

b/ x^15 = x

mak ta có 0 mũ bao nhiêu cx = 0 => x=0

c/ ( 2x + 1 )^3  = 125

mak 5^3 =125

ta có 2x + 1 = 5

    2x = 5 - 1

     2x = 4

       x = 4 : 2 = 2

a ) 2 ^x  . 4 = 128

    2 ^x          = 32

    2 ^x          = 2 ⁵

=> x = 5

b ) x ¹⁵ = x

=> x = 0 hoặc x = 1

c ) ( 2x + 1 ) ³ = 125

     ( 2x + 1 ) ³ = 5 ³

=> 2x + 1 = 5

     2x       = 4

       x       = 2

d ) ( x - 5 ) ⁴ = ( x - 5 ) ⁶

=> x - 5 = 0 hoặc x - 5 = 1

         x  = 5              x  = 6

Vậy x = 5 hoặc x = 6

a) x=5

b) x=1 hoặc x=0 hoặc x=-1

c) x=2

d) x=5 hoặc x=-4 hoặc x=6

a) \(2^x.4=128\)

\(2^x.2^2=2^7\)

\(2^x=2^7:2^2\)

\(2^x=2^5\)

\(x=5\)

a)\(2^x.4=128\Leftrightarrow2^x=32\Leftrightarrow2^x=2^5\Rightarrow x=5\)

b)\(\left(2x+1\right)=125\Leftrightarrow2x=126\Leftrightarrow x=13\)

c)\(x^{15}=x\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=0\end{cases}}\)

d) \(\left(x-5\right)^4=\left(x-5\right)^5\Leftrightarrow\orbr{\begin{cases}x-5=1\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}\)

a, 

\(2^x=32\)     

\(2^x=2^5\)   

\(\Rightarrow x=5\)   

b, 

2x = 124 

x = 62 

c, 

\(x^{15}-x=0\)   

\(x\left(x^{14}-1\right)=0\)   

\(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\)   

\(\orbr{\begin{cases}x=0\\x^{14}=1\end{cases}}\)  

\(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)               

d, 

\(0=\left(x-5\right)^5-\left(x-5\right)^4\)   

\(\left(x-5\right)^4\left(x-5-1\right)=0\)   

\(\orbr{\begin{cases}\left(x-5\right)^4=0\\x-6=0\end{cases}}\)      

\(\orbr{\begin{cases}x=5\\x=6\end{cases}}\)

\(2^x.4=128\)

\(2^x=128:4\)

\(2^x=32\)

\(\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)

\(x^{15}=x\Leftrightarrow x\in\left\{-1;0;1\right\}\)

\(\left(2x+1\right)^3=125\)

\(\Leftrightarrow\left(2x+1\right)^3=5^3\)

\(\Leftrightarrow2x+1=5\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

\(\left(x-5\right)^6=\left(x-5\right)^4\)

\(\Leftrightarrow\hept{\begin{cases}x-5=-1\\x-5=0\\x-5=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\x=5\\x=6\end{cases}}\)

\(\text{Vậy:}\)\(x\in\left\{4;5;6\right\}\)

\(2^x.4=128\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5.\)

\(x^{15}=x\Rightarrow\orbr{\begin{cases}x=\pm1\\x=0\end{cases}}\)

       \(\left(2x+1\right)^3=125\)

<=>  \(\left(2x+1\right)^3=5^3\)

<=>   \(2x+1=5\)

<=>   \(x=2\)

        \(\left(x-5\right)^6=\left(x-5\right)^4\)

<=>   \(\left(x-5\right)^6-\left(x-5\right)^4=0\)

<=>    \(\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)

<=>     \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

<=>      \(\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)

Giải ra được x = 5 ; x = 6 ; x = 4 .