x^5=x      

=>x=0 hặc 1      

(2x+1)^3=125. 

(2x+1)³=5³

=>2x+1=5

2x=5-1

2x=4

x=4:2

x=2

A =3+3²+3³+...+3¹⁰⁰

3A=3(3+3²+3³+...+3¹⁰⁰)

3A=3²+3³+3⁴+...+3¹⁰¹

3A-A=(3²+3³+3⁴+...+3¹⁰¹)-(3+3²+3³+...+3¹⁰⁰)

2A=3¹⁰¹-3

=>2A+3=3^n

3¹⁰¹-3+3=3ⁿ

3¹⁰¹=3ⁿ

vậy n=101

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, \(x^{15}=x\)

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

b, \(\left(2x+1\right)^3=125\)

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

\(\Rightarrow\) \(2x+1=5\)

\(\Rightarrow\) \(2x=5-1\)

\(\Rightarrow\) \(2x=4\)

\(\Rightarrow\) \(x=4:2\)

\(\Rightarrow\) \(x=2\)

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

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-5=1\end{cases}}\)

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

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) \(2^x\cdot4=128\)

\(2^x=32\)

\(x=5\)

b) \(x^{15}=x\)

\(\Leftrightarrow x\in\left\{0;1\right\}\)

c) \(\left(2x+1\right)^3=125\)

\(\Rightarrow2x+1=5\)

\(2x=4\)

\(x=2\)

a)x=5

b)x=0 hoặc x=1

c)x=2

d)x=5 hoặc x=6

a, x= 5  
b, x= 1
c, x = 2
d, x =6