Vì \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow2x-15=\orbr{\begin{cases}0\\1\end{cases}}\)vì chỉ có \(0^5=0^3;1^5=1^3\)
\(\Rightarrow2x=\orbr{\begin{cases}15\\16\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}\frac{15}{2}\\8\end{cases}}\)
\(\left(2x-15\right)^5=\left(2x-15\right)^3.\)
\(\Rightarrow\hept{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow\hept{\begin{cases}2x=15\\2x=16\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}}\)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)
\(\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)
\(\Rightarrow\left(2x-15\right)^3=0\)hoac \(\orbr{\begin{cases}2x-16=0\\2x-14=0\end{cases}}\)
\(\Rightarrow2x-15=0\)hoac \(\orbr{\begin{cases}2x=16\\2x=14\end{cases}}\)
\(\Rightarrow2x=15\) hoac \(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)
\(\Rightarrow x=\frac{15}{2}\)hoac \(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)
vay \(x=\frac{15}{2}\)hoac \(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)
Ta có : (2x - 15)5 = (2x-15)3
=> (2x - 15)5 - (2x - 15)3 = 0
=> (2x - 15)3[(2x - 15)2 - 1] = 0
\(\Leftrightarrow\hept{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x-15\right)=0\\\left(2x-15\right)^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x=15\\\left(2x-15\right)=-1;1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\2x=14;16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=7,5\\x=7;8\end{cases}}\)
Vậy x = {7;7,5;8}
(2x-15)5 = (2x-15)3 =1
2x-15 =1
2x =1+15
2x =16
x =16:2
x =8
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3.[\left(2x-15\right)^2-1]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{15}{2}\\\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}}\end{cases}}\)
(2x-15)5=(2x-3)3
=> (2x-15)5-(2x-3)3=0
=>(2x-15)3.((2x-15)2-1)=0
=>(2x-15)3=0 hoặc(2x-15)2-1=0
=>x=7,5 hoặc x=8 hoặc x=7
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=15\\2x=16\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}\)
Ta có : ( 2.x - 15 )5 = ( 2.x - 15 ) 3
<=> ( 2.x -15 )5 - ( 2.x - 15 ) 3 = 0
<=> ( 2.x - 15 )3 . [ ( 2.x - 15 )2 - 1 ] = 0
<=> \(\orbr{\begin{cases}\left(2.x-15\right)^3=0\\\left[\left(2.x-15\right)^2-1\right]=0\end{cases}}\)
<=> \(\orbr{\begin{cases}2.x-15=0\\2.x-15=1\end{cases}}\)
<=> \(\orbr{\begin{cases}2.x=15\\2.x=16\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}\)
Vậy x = { 8 ; 7,5 }
