Помогите с алгеброй!!!! Нужен только один пример!!! Срочно!!! 36.6 под а. дам 20 баллов.

[latex]\frac{8}{16x^2-9}- \frac{8}{16x^2-24x+9} = \frac{1}{4x^2+3x}\\ \frac{8}{(4x)^2-3^2}- \frac{8}{(4x)^2-2*4x*3+3^2} = \frac{1}{x(4x+3)}\\ \frac{8}{(4x-3)(4x+3)}- \frac{8}{(4x-3)^2} = \frac{1}{x(4x+3)}\\ \frac{8x(4x-3)}{x(4x-3)^2(4x+3)}- \frac{8x(4x+3)}{x(4x-3)^2(4x+3)} = \frac{(4x-3)^2}{x(4x-3)^2(4x+3)}\\ \frac{8x(4x-3)}{x(4x-3)^2(4x+3)}- \frac{8x(4x+3)}{x(4x-3)^2(4x+3)} - \frac{(4x-3)^2}{x(4x-3)^2(4x+3)}=0\\[/latex] [latex]\frac{8x(4x-3)-8x(4x+3) -(4x-3)^2}{x(4x-3)^2(4x+3)}=0\\ \frac{32x^2-24x-32x^2-24x-((4x)^2-2*4x*3+3^2)}{x(4x-3)^2(4x+3)}=0\\ \frac{-48x-(16x^2-24x+9)}{x(4x-3)^2(4x+3)}=0\\ \frac{-48x-16x^2+24x-9}{x(4x-3)^2(4x+3)}=0\\ \frac{-24x-16x^2-9}{x(4x-3)^2(4x+3)}=0\\ ODZ x \neq 0 \\ \\4x-3 \neq 0\\ 4x \neq 3\\ x \neq 3/4\\ x \neq 0.75 \\ \\4x+3 \neq 0\\4x \neq -3\\ x \neq -3/4\\ x \neq -0.75 \\[/latex] -16x²-24x-9=0 D = (-24)² - 4·(-16)·(-9) = 576 - 576 = 0 x = 24/(2·(-16))=-24/32=-3/4 = -0.75 Нет решения