a)

 \(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)

b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)

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

Chọn D.

(x − 5)(x + 3) = x(x + 3) – 5( x + 3) =  x 2  + 3x - 5x - 15 =  x 2 − 2x − 15

a) \(\left(2x-1\right)+\frac{3}{15}=\frac{3}{2}\)

\(\Rightarrow2x-1=\frac{3}{2}-\frac{3}{15}=\frac{13}{10}\)

\(\Rightarrow2x=\frac{13}{10}+1=\frac{23}{10}\)

\(\Rightarrow x=\frac{23}{20}\)

b) \(x+\frac{46}{15}=1,5\)

\(\Rightarrow x+\frac{46}{15}=\frac{3}{2}\)

\(\Rightarrow x=\frac{3}{2}-\frac{46}{15}\)

\(\Rightarrow x=\frac{-47}{30}\)

c) \(\left(-2x+1\right)+\frac{3}{15}=\frac{5}{3}\)

\(\Rightarrow-2x+1=\frac{5}{3}-\frac{3}{15}=\frac{22}{15}\)

\(\Rightarrow-2x=\frac{7}{15}\Rightarrow x=\frac{-7}{30}\)

\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=\pm1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\) 

Vậy.........

Ta có: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Leftrightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-15-1\right)\left(2x-15+1\right)=0\)

\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{15}{2};8;7\right\}\)

Tham khảo : 

`(2x - 15)^5 = (2x - 15)^3`

`=> (2x - 15)^5 : (2x - 15)^3 = 1`

`=> (2x - 15)^2 = 1`

`=> (2x - 15)^2 = 1^2`

`=>` $\left[\begin{matrix} 2x-15=1\\ 2x-15=-1\end{matrix}\right.$

`=>` $\left[\begin{matrix} 2x=1 + 15\\ 2x=-1 + 15\end{matrix}\right.$

`=>` $\left[\begin{matrix} 2x=16\\ 2x=14\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=8\\ x=7\end{matrix}\right.$

`=> x in {7;8}`

`(2x-15)^5 =(2x-15)^3`

`=>(2x-15)^5 -(2x-15)^3=0`

`=> (2x-15)^3 [(2x-15)^2 -1]=0`

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=1\\2x-15=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(=>\left[{}\begin{matrix}2x-15=0\\2x-15=1\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}2x=15\\2x=16\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\end{matrix}\right.\)

\(=>x\in\left\{\dfrac{15}{2};8\right\}\)