a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6\right)^2=\left(1-x\right)^2\\-3< =x< =1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6+x-1\right)\left(2x+6+1-x\right)=0\\-3< =x< =1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x+5\right)\left(x+7\right)=0\\-3< =x< =1\end{matrix}\right.\Leftrightarrow x=-\dfrac{5}{3}\)
b: \(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=2x\)
\(\Leftrightarrow4\sqrt{x-3}=2x\)
\(\Leftrightarrow2\sqrt{x-3}=x\)
\(\Leftrightarrow\sqrt{4x-12}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2=4x-12\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
