Số nghiệm của phương trình ${{\log }_{2}}(x+2)+{{\log...

Điều kiện: $-2<x\ne 5$
${{\log }_{2}}(x+2)+{{\log }_{4}}{{(x-5)}^{2}}+{{\log }_{\dfrac{1}{2}}}8=0\Leftrightarrow {{\log }_{2}}(x+2)+{{\log }_{2}}\left| x-5 \right|={{\log }_{2}}8$
$\Leftrightarrow {{\log }_{2}}\left[ \left( x+2 \right)\left| x-5 \right| \right]={{\log }_{2}}8\Leftrightarrow \left( x+2 \right)\left| x-5 \right|=8$
$\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& x>5 \\
& \left( x+2 \right)\left( x-5 \right)=8 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& -2<x<5 \\
& \left( x+2 \right)\left( 5-x \right)=8 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& x>5 \\
& {{x}^{2}}-3x-18=0 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& -2<x<5 \\
& -{{x}^{2}}+3x+2=0 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow $ $ \left[ \begin{aligned}
& \left\{ \begin{aligned}
& x>5 \\
& x=6\vee x=-3 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& -2<x<5 \\
& x=\dfrac{3+\sqrt{17}}{2}\vee x=\dfrac{3-\sqrt{17}}{2} \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=6 \\
& x=\dfrac{3+\sqrt{17}}{2} \\
& x=\dfrac{3-\sqrt{17}}{2} \\
\end{aligned} \right.$