![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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}`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(=>\left[{}\begin{matrix}2x-15=0\\2x-15=1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}2x=15\\2x=16\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\end{matrix}\right.\)
\(=>x\in\left\{\dfrac{15}{2};8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)
\(\Leftrightarrow x=\frac{15}{2};8;7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x-15)5=(2x-15)3
=>(2x-15)5-(2x-15)3=0
=>(2x-15)3.(2x-15)2-(2x-15)3.1=0
=>(2x-15)3.((2x-15)2-1)=0
=>(2x-15)3=0=>2x-15=0=>2x=15=>x=15/2
hoặc (2x-15)2-1=0=>(2x-15)2=0=>2x-15=1,-1=>2x=16,14=>x=8,7
Vậy x=15/2,8,7.
ta co ( 2x-15)5= (2x-15)3
=> (2x-15)5-(2x-15)3=0
=> (2x-15)3 .{(2x-15)2-1}=0
=> (2x-15)3=0 hoac (2x-15)2-1=0
=> 2x=15 hoac (2x-15)2=1
=> x=15/2 hoac 2x-15=-1;1
=> x=15/2 hoac 2x= 14;16 => x=7;8
vay x=15/2;7;8
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=\pm1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
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: \(x\in\left\{\dfrac{15}{2};8;7\right\}\)