A=7 B=1 OR =0 C=2 D=4.5.6

a: \(2^{x}\cdot4=128\)

=>\(2^{x}=\frac{128}{4}=32=2^5\)

=>x=5

b: \(x^{15}=x\)

=>\(x^{15}-x=0\)

=>\(x\left(x^{14}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{14}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{14}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)

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

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

=>2x+1=5

=>2x=5-1=4

=>\(x=\frac42=2\)

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

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

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

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

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

=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\\ x=4\end{array}\right.\)

Vì 125 ⋮ x, 100 ⋮ x, 150 ⋮ x nên x ∈ UC(125,100,150)

mà x là lớn nhất nên x = UCLN(125,100,150) = 25

`@` `\text {Ans}`

`\downarrow`

`2^x * 4 = 128`

`=> 2^x = 128 * 4`

`=> 2^x = 512`

`=> 2^x = 2^9`

`=> x = 9`

Vậy, `x = 9`

`x^15 = x`

`=> x^15 - x = 0`

`=> x(x^14 - 1) = 0`

`=>` TH1: `x = 0`

`TH2: x^14 - 1 = 0`

`=> x^14 = 1`

`=> x = 1`

Vậy, `x \in {0; 1}`

`(2x+1)^3 = 125`

`=> (2x+1)^3 = 5^3`

`=> 2x + 1 = 5`

`=> 2x = 5 - 1`

`=> 2x =4`

`=> x = 4 \div 2`

`=> x = 2`

Vậy,` x = 2.`

`(x - 5)^4 = (x-5)^6`

`=> (x-5)^4 - (x-5)^6 = 0`

`=> (x-5)^4 * [ 1 - (x-5)^2] = 0`

`=> - (x-6)(x-5)^4(x-4) = 0`

`TH1: (x - 5)^4 = 0`

`=> x - 5 = 0`

`=> x = 0 +5`

`=> x = 5`

`TH2: x - 6=0`

`=> x=6`

`TH3: x-4=0`

`=> x = 4`

Vậy, `x \in {4; 5; 6}`

a: =>2^x=32

=>x=5

b: =>x^15-x=0

=>x(x^14-1)=0

=>x=0; x=1;x=-1

c: =>2x+1=5

=>2x=4

=>x=2

d: =>(x-5)^4[(x-5)^2-1]=0

=>(x-5)(x-4)(x-6)=0

=>x=5;x=4;x=6

UCLN(100;125;150) = 25

UC(100;125;150) = U(25) = {1;5;25}

Vì x < 10 nên x ∈ {1;5}

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 = 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 }