Đáp án + giải thích các bước giải:

\begin{array}{1}TXĐ: \begin{cases} 2x-3\ge0 \\ 3x-2\ge0 \end{cases}\rightarrow \begin{cases} x\ge\dfrac{3}{2} \\ x\ge\dfrac{2}{3} \end{cases}\rightarrow D=[\dfrac{3}{2};+\infty) \\(5x-4)\sqrt{2x-3}-(4x-5)\sqrt{3x-2}=2 \\ \rightarrow (5x-4)\sqrt{2x-3}=2+(4x-5)\sqrt{3x-2} \\ \rightarrow (5x-4)^2(2x-3)=[(2+(4x-5)\sqrt{3x-2}]^2 \\ \rightarrow 50x^3-155x^2+152x-48=48x^3-152x^2+155x-46+4(4x-5)\sqrt{3x-2} \\ \rightarrow 2x^3-3x^2-3x-2-4(4x-5)\sqrt{3x-2}=0 \\ \rightarrow 2x^3-3x^2-3x-2-4(4x-5)\sqrt{3x-2}+(4x-5)(3x-2)-(4x-5)(3x-2)=0 \\ \rightarrow 2x^3-3x^2-3x-2 +(4x-5)\sqrt{3x-2}(-4+\sqrt{3x-2})-(4x-5)(3x-2) \\ \rightarrow 2x^3-15x^2+20x-12 +(4x-5)\sqrt{3x-2}(-4+\sqrt{3x-2})=0 \\ \rightarrow (x-6)(2x^2-3x+2)+\dfrac{(4x-5)\sqrt{3x-2}(3x-2-16)}{\sqrt{3x-2}+4}=0 \\ \rightarrow (x-6)(2x^3-3x+2)+\dfrac{3(x-6)(4x-5)\sqrt{3x-2}}{\sqrt{3x-2}+4}=0 \\ \rightarrow (x-6)[\dfrac{3(4x-5)\sqrt{3x-2}}{\sqrt{3x-2}+4}+2x^2-3x+2]=0 \\ \text{Với } x\ge \dfrac{3}{2} \text{ dễ thấy } \dfrac{3(4x-5)\sqrt{3x-2}}{\sqrt{3x-2}+4}+2x^2-3x+2>0 \\ \rightarrow x-6=0 \\ \rightarrow x=6 (TM) \\ \end{array} ` \text{Vậy }S={6}`