a.

2^x . 4 = 128

2^x = 128 : 4

2^x = 32

2^x = 2⁵

x = 5

b.

x¹⁵ = x

Vậy x = 0 hoặc x = 1 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

TH1:

x - 5 = 0

x = 5

TH2:

x - 5 = -1

x = -1 + 5

x = 4

TH2:

x - 5 = 1

x = 1 + 5 

x = 6

Vậy x = 5 hoặc x = 4 hoặc x = 6

Chúc bạn học tốt ^^ 

a) \(2^x.4=128\)

=> \(2^x=32\) => \(2^x=2^5\) => x = 5

b) \(x^{15}=x\) => x = 1 hoặc x = 0

c) \(\left(2x+1\right)^3=125\)

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

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

=> x - 5 = 0 hoặc x - 5 = 1

=> x = 5 hoặc x= 6

Chúc bạn làm bài tốt

thank you very much

\(a,2^x.4=128\)

\(2^x\) \(=128:4\)

\(2^x\) \(=32\)

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

\(x\) \(=5\)

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

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

Vậy \(\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

\(c,(2x+1)^3=125\)

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

\(2x+1\) \(=5\)

\(2x\) \(=5-1\)

\(2x\) \(=4\)

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

\(x\) \(=2\)

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

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=-1\\x-5=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0+5\\x=\left(-1\right)+5\\x=1+5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)