$DK: \ \left[ \begin{array}{l}x \le - 3 \\ x \ge 1 \\ \end{array} \right. \\ pt\iff \sqrt {2{x^2} + 11x + 15} - \left( {x + \dfrac{9}{2}} \right) + \sqrt {{x^2} + 2x - 3} - \dfrac{3}{2}=0 \\ \iff \dfrac{{2{x^2} + 11x + 15 - {{\left( {x + \dfrac{9}{2}} \right)}^2}}}{{\sqrt {2{x^2} + 11x + 15} + x + \dfrac{9}{2}}} + \dfrac{{{x^2} + 2x - 3 - \dfrac{9}{4}}}{{\sqrt {{x^2} + 2x - 3} + \dfrac{3}{2}}}=0 \\ \iff \dfrac{{\left( {2x + 7} \right)\left( {2x - 3} \right)}}{{4\sqrt {\left( {x + 3} \right)\left( {2x + 5} \right)} + 4x + 18}} + \dfrac{{\left( {2x + 7} \right)\left( {2x - 3} \right)}}{{4\sqrt {\left( {x - 1} \right)\left( {x + 3} \right)} + 6}}=0 \\ \iff \left({2x+7} \right)\left( {2x-3} \right)\left({\dfrac{1}{{4\sqrt{\left( {x + 3} \right)\left( {2x+5} \right)}+4x+18}} + \dfrac{1}{{4\sqrt {\left( {x - 1} \right)\left( {x + 3} \right)} + 6}}} \right)=0 \\ \iff \left( {2x + 7} \right)\left( {2x - 3} \right) =0 \\ \iff \left[ \begin{array}{l}x =\dfrac{3}{2} \\ x = \dfrac{{ - 7}}{2} \\ \end{array} \right.$
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