Bài giải

PT1: $2{{\log }_{5}}(3x-1)+1={{\log }_{\sqrt[3]{5}}}(2x+1)$

$\Leftrightarrow \left\{ \begin{align}& 3x-1>0 \\ & 2x+1>0 \\ & 2{{\log }_{5}}(3x-1)+1={{\log }_{\sqrt[3]{5}}}(2x+1) \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x>\frac{1}{3} \\ & {{\log }_{5}}{{(3x-1)}^{2}}+{{\log }_{5}}5=3{{\log }_{5}}(2x+1) \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>\frac{1}{3} \\ & {{\log }_{5}}5{{(3x-1)}^{2}}={{\log }_{5}}{{(2x+1)}^{3}} \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x>\frac{1}{3} \\ & 5{{(3x-1)}^{2}}={{(2x+1)}^{3}} \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>\frac{1}{3} \\ & 5(9{{x}^{2}}-6x+1)=8{{x}^{3}}+12{{x}^{2}}+6x+1 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x>\frac{1}{3} \\ & 8{{x}^{3}}-33{{x}^{2}}+36x-4=0 \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>\frac{1}{3} \\ & x=\frac{1}{8}\vee x=2 \\ \end{align} \right.\Rightarrow {{x}_{1}}=2$

PT2: ${{\log }_{2}}({{x}^{2}}-2x-8)=1-{{\log }_{\frac{1}{2}}}(x+2)$

$\Leftrightarrow \left\{ \begin{align}& {{x}^{2}}-2x-8>0 \\ & x+2>0 \\ & {{\log }_{2}}({{x}^{2}}-2x-8)=1-{{\log }_{\frac{1}{2}}}(x+2) \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x<-2\vee x>4 \\& x>-2 \\ & {{\log }_{2}}({{x}^{2}}-2x-8)=1+{{\log }_{2}}(x+2) \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>4 \\& {{\log }_{2}}({{x}^{2}}-2x-8)={{\log }_{2}}2(x+2) \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>4 \\ & {{x}^{2}}-2x-8=2(x+2) \\\end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>4 \\& {{x}^{2}}-4x-12=0 \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x>4 \\ & x=-2\vee x=6 \\ \end{align} \right.\Rightarrow {{x}_{2}}=6$ 

Vậy ${{x}_{1}}+{{x}_{2}}=2+6=8$.