Кто шарит , решите 3 Даю 30б

[latex] \frac{2}{x-5} - \frac{3x-6}{6-2x} = \frac{3}{(x-3)(x-1)}\; ,\; \; \; ODZ:\; \; x\ne 5,\; x\ne 3,\; x\ne 1\\\\ \frac{2}{x-5} +\frac{3x-6}{2(x-3)} - \frac{3}{(x-3)(x-1)} =0\\\\ \frac{4(x-3)(x-1)+(3x-6)(x-5)(x-1)-6(x-5)}{2(x-5)(x-3)(x-1)} =0\\\\ \frac{4(x^2-4x+3)+(3x^2-21x+30)(x-1)-6x+30}{2(x-5)(x-3)(x-1)} =0\\\\ \frac{3x^3-20x^2+29x+12}{2(x-5)(x-3)(x-1)} =0[/latex] [latex]3x^3-20x^2+29x+12=0\\\\x=3:\; \; 3\cdot 3^3-20\cdot 3^2+29\cdot 3+12=0\\\\x=4:\; \; 2\cdot 4^3-20\cdot 4^2+29\cdot 4+12=0\; \; \to \\\\ 3x^3-20x^2+29x+12=(x-3)(x-4)(3x+1)\\\\x=3\notin ODZ\\\\x=4\; ,\; \; x=-\frac{1}{3}\; \; -\; \; otvet[/latex]

[latex] \frac{2}{x-5}- \frac{3x-6}{6-2x}= \frac{3}{(x-3)(x-1)}; \\ x-5=t; \\ \frac{2}{t}+ \frac{3(t+3)}{2(t+2)}= \frac{3}{(t+2)(t+4)}; \\ \frac{3t^2+13t+8}{2t(t+2)}= \frac{3}{(t+2)(t+4)}; \\ (3t^2+13t+8)*(t+2)(t+4)=6t(t+2); \\ (t+2)(3t^3+25t^2+54t+32)=0; \\ (t+2)((t+1)(3t+16)(t+2)=0; \\ t+2=0; t=-2; x-5=-2; x=3; \\ t+1=0; t=-1;x-5=-1; x=4; \\ 3t+16=0; t=- \frac{16}{3}; x-5=- \frac{16}{3}; x=- \frac{1}{3}. [/latex] ОДЗ: x≠5, x≠3, x≠1. Ответ: -1/3; 4.