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

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

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`

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow x=\frac{15}{2};8;7\)

Ta có : 

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

\(\Leftrightarrow\)\(\left(2x-15\right).1=\left(2x-15\right)\left(2x-15\right)\)

\(\Leftrightarrow\)\(2x-15=1\)

\(\Leftrightarrow\)\(2x=16\)

\(\Leftrightarrow\)\(x=\frac{16}{2}\)

\(\Leftrightarrow\)\(x=8\)

Vậy \(x=8\)

Chúc bạn học tốt ~ 

Ta có : 

(2x−15)=(2x−15)2

⇔(2x−15).1=(2x−15)(2x−15)

⇔2x−15=1

⇔2x=16

⇔x=162 

⇔x=8

Vậy x=8

Chúc bạn học tốt 

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

\(\Rightarrow\left(2x-15\right)=\left(2x-15\right).\left(2x-15\right)\)

\(\Rightarrow\left(2x-15\right).1=\left(2x-15\right).\left(2x-15\right)\)

\(\Rightarrow2x-15=1\)

\(\Rightarrow2x=16\)

\(\Rightarrow x=8\)

Nguồn: Phùng Minh Quân

Mình chỉ làm hơi khác tí thôi nha, đầy đủ hơn so với cách lớp 7 của anh Quân.

Chọn D.

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