(2x-15)5=(2x-15)3
(2x-15)5=(2x-15)3
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=\pm1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
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: \(x\in\left\{\dfrac{15}{2};8;7\right\}\)
(2x-15)^5=(2x-15)^3
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
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(=>\left[{}\begin{matrix}2x-15=0\\2x-15=1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}2x=15\\2x=16\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\end{matrix}\right.\)
\(=>x\in\left\{\dfrac{15}{2};8\right\}\)
(2x-15)5=(2x-15)3
`(2x-15)^5 =(2x-15)^3`
`=>(2x-15)^5 -(2x-15)^3=0`
`=> (2x-15)^3 [(2x-15)^2 -1]=0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=1\\2x-15=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
( 2x - 15 ) ^ 5 = ( 2x -15 ) ^ 3
(2x-15)^5=(2x-15)^3
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)
\(\Leftrightarrow x=\frac{15}{2};8;7\)
Tim x , : a)(2x-15)^5=(2x-15)^3
Tìm x , biết:
a) (2x-15)5 = (2x-15)3
b) (7x-11)3 = (-3)2 . 15 + 208
Giúp mình nhanh nhé
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) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)
Tình hợp lý nếu có thể :
-5/2x 2/11+-5/7x 9/11+15/7
Tìm x biết:
(2x-15)mũ 5=(2x-15)mũ 3
25/7nha
(2x-15)5=(2x-15)3