a: \(x^2-2\left(x+3\right)=x-6\)
=>\(x^2-2x-6-x+6=0\)
=>\(x^2-3x=0\)
=>x(x-3)=0
=>\(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
b: \(1\dfrac{3}{5}+\left(\dfrac{\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{37}}{\dfrac{5}{7}+\dfrac{5}{17}+\dfrac{5}{37}}\right)x=\dfrac{16}{5}\)
=>\(\dfrac{8}{5}+\left(\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{37}\right)}{5\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{37}\right)}\right)x=\dfrac{16}{5}\)
=>\(\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\)
=>2x+8=16
=>2x=8
=>x=4
c: \(\left(2x-1\right)^4=16\)
=>\(\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: \(\left(2x+1\right)^6=\left(2x+1\right)^4\)
=>\(\left(2x+1\right)^6-\left(2x+1\right)^4=0\)
=>\(\left(2x+1\right)^4\left[\left(2x+1\right)^2-1\right]=0\)
=>\(\left[{}\begin{matrix}\left(2x+1\right)^4=0\\\left(2x+1\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\2x+1=1\\2x+1=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=0\\x=-1\end{matrix}\right.\)