Phương trình $\dfrac{1}{2}{{\log }_{\sqrt{3}}}\left(x+3 \right)+\dfrac{1}{2}{{\log }_{9}}{{\left(x-1 \right)}^{4}}=2{{\log }_{9}}\left(4x \right)$...

Điều kiện: $\left\{ \begin{aligned}
& x+3>0 \\
& x-1\ne 0 \\
& 4x>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>-3 \\
& x\ne 1 \\
& x>0 \\
\end{aligned} \right.\Leftrightarrow 0<x\ne 1.$
Ta có: $\dfrac{1}{2}{{\log }_{\sqrt{3}}}\left(x+3 \right)+\dfrac{1}{2}{{\log }_{9}}{{\left(x-1 \right)}^{4}}=2{{\log }_{9}}\left(4x \right)\Leftrightarrow {{\log }_{3}}\left(x+3 \right)+{{\log }_{3}}\left| x-1 \right|={{\log }_{3}}\left(4x \right)$
$\Leftrightarrow {{\log }_{3}}\left(x+3 \right)\left| x-1 \right|={{\log }_{3}}\left(4x \right)\Leftrightarrow \left(x+3 \right)\left| x-1 \right|=4x\left(* \right).$
Trường hợp 1: Nếu $x>1$ thì $\left(* \right)\Leftrightarrow \left(x+3 \right)\left(x-1 \right)=4x\Leftrightarrow {{x}^{2}}-2x-3=0\Leftrightarrow \left[ \begin{aligned}
& x=-1\left(loa\ddot{i}i \right) \\
& x=3 \\
\end{aligned} \right.$
Trường hợp 2: Nếu $0<x<1$ thì
$\left(* \right)\Leftrightarrow \left(x+3 \right)\left(1-x \right)=4x\Leftrightarrow {{x}^{2}}+6x-3=0\Leftrightarrow \left[ \begin{aligned}
& x=-3-2\sqrt{3}\left(loa\ddot{i}i \right) \\
& x=-3+2\sqrt{3} \\
\end{aligned} \right.$
Kết luận: Phương trình đã cho có 2 nghiệm thực.