ĐK \(\left\{{}\begin{matrix}x>3\\x\ge-1\end{matrix}\right.\)
\(\left(x-3\right)\left(x+1\right)+4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{\left(x-3\right)\left(x+1\right)}=-1\left(vl\right)\\\sqrt{\left(x-3\right)\left(x+1\right)}=-3\left(vl\right)\end{matrix}\right.\)
vậy pt vô nghiệm