Найдём ограничения, знаменатель не должен равнятся нулю:

[tex]2 {x}^{2} + x - 3 = 0 \\ a = 2 \\ b =1 \\ c = - 3\\ D = {b}^{2} - 4ac = 1 {}^{2} - 4 \times 2 \times ( - 3) = \\ = 1 + 24 = 25 \\ x _{1}= \frac{ - 1 - 5}{2 \times 2} = - \frac{6}{4} = - 1.5 \\ x_{2} = \frac{ - 1 + 5}{2 \times 2} = \frac{4}{4} = 1[/tex]

[tex]2 {x}^{2} - 3x - 9 = 0 \\ a = 2 \\ b = - 3\\ c = - 9 \\ D = {b}^{2} - 4ac = ( - 3) {}^{2} - 4 \times 2 \times ( - 9) = \\ = 9 + 72 = 81 \\ x_{1} = \frac{3 - 9}{2 \times 2} = - \frac{6}{4} = - 1.5 \\ x _{2}= \frac{3 + 9}{2 \times 2} = \frac{12}{4} = 3[/tex]

Получаем:

[tex]x\neq - 1.5 \\ x\neq 1 \\ x\neq3[/tex]

[tex] \frac{2 {x}^{2} }{2 {x}^{2} + x - 3 } - \frac{8}{2 {x}^{2} - 3x - 9} - 1 = 0 \\ \frac{2 {x}^{2} }{2(x + 1.5)(x - 1)} - \frac{8}{2(x + 1.5)(x - 3)} - 1 = 0 \\ \frac{2 {x}^{2}( x- 3) - 8(x - 1) - (2x + 3)(x - 1)(x - 3)}{2(x + 1.5)(x - 1)(x - 3)} = 0 \\ 2 {x}^{2}( x- 3) - 8(x - 1) - (2x + 3)(x - 1)(x - 3) = 0 \\ 2 {x}^{3} - 6 {x}^{2} - 8x + 8 - (2x + 3)( {x}^{2} - 3x - x + 3) = 0 \\ 2 {x}^{3} - 6 {x}^{2} - 8x + 8 - (2 {x}^{3} - 8 {x}^{2} + 6x + 3 {x}^{2} - 12x + 9) = 0 \\ 2 {x}^{3} - 6 {x}^{2} - 8x + 8 - 2 {x}^{3} + 5 {x}^{2} + 6x - 9 = 0 \\ - {x}^{2} - 2x - 1 = 0 \\ {x}^{2} + 2x + 1 = 0 \\ (x + 1) { }^{2} = 0 \\ x + 1 = 0 \\ x = - 1[/tex]

Ответ: х = - 1