\(\sqrt{3x-2}+\sqrt{3+x}=\sqrt{5x+4}\)
→ \(\left(\sqrt{3x-2}+\sqrt{3+x}\right)^2=\left(\sqrt{5x+4}\right)^2\)
→ \(3x-2+3+x+2\sqrt{\left(2x-2\right)\left(3+x\right)}=5x+4\)
➝ \(4x+3+2\sqrt{6x+2x^2-6-2x}=5x+4\)
→ \(2\sqrt{2x^2+4x-6}=5x+4-4x-3\)
→ \(2\sqrt{2x^2+4x-6}=x+1\)
→ \(\left(2\sqrt{2x^2+4x-6}\right)^2=\left(x+1\right)^2\)
→ \(4\left(2x^2+4x-6\right)=x^2+2x+1\)
→ \(8x^2+16x-24=x^2+2x+1\)
→ \(8x^2+16x-24-x^2-2x-1=0\)
→ \(7x^2+14x-25=0\)
→ \(x_1=\frac{-7+4\sqrt{14}}{7}\)
\(x_2=\frac{-7-4\sqrt{14}}{7}\)
\( \sqrt {3x - 2} + \sqrt {3 + x} = \sqrt {5x + 4} \left( {x \ge \dfrac{2}{{3}}} \right)\\ \Leftrightarrow 3x - 2 + 2\sqrt {\left( {3x - 2} \right)\left( {3 + x} \right)} + 3 + x = 5x + 4\\ \Leftrightarrow 2\sqrt {7x + 3{x^2} - 6} = x + 3\\ \Leftrightarrow 4\left( {7x + 3{x^2} - 6} \right) = {x^2} + 6x + 9\\ \Leftrightarrow 28x + 12{x^2} - 24 = {x^2} + 6x + 9\\ \Leftrightarrow {x^2} + 2x - 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 1\left( {tm} \right)\\ x = - 3\left( {ktm} \right) \end{array} \right. \)