a. \(x^3=x\Leftrightarrow x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
b. \(x^{2019}=x^{2018}\Leftrightarrow x^{2019}-x^{2018}=0\)
\(\Leftrightarrow x^{2018}\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^{2018}=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
c. \(3^{x+1}+3^x=108\)
\(\Leftrightarrow3^x\left(3+1\right)=108\Leftrightarrow3^x.4=108\)
\(\Leftrightarrow3^x=27\Leftrightarrow x=3\)
d. \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Leftrightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Leftrightarrow\left(x-5\right)^4.\left[1-\left(x-5\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\\\\left(x-5\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
a)x^3 = x
KQ:
x = +1
x = 0
