Рационал уравнения помогитеее

3)
$\frac{x}{x+10} = \frac{1}{x-8} \\ \\ \frac{x}{x+10} - \frac{1}{x-8} =0 \\ \\ \frac{x(x-8)-(x+10)}{(x+10)(x-8)} =0 \\ \\ \frac{x^2-8x-x-10}{(x+10)(x-8)} =0 \\ \\ \frac{x^2-9x-10}{(x+10)(x-8)} =0 \\ \\ \frac{(x-10)(x+1)}{(x+10)(x-8)} =0 \\ \\ \left \{ {{(x-10)(x+1)=0} \atop {(x+10)(x-8) \neq 0}} \right. \\ \\ 1) x-10 = 0; x_1=10 \\ \\ 2)x+1=0; x_2=-1$

4)
$\frac{1}{3x-1} = \frac{x}{27-x} \\ \\ \frac{1}{3x-1} - \frac{x}{27-x} =0 \\ \\ \frac{27-x-x(3x-1)}{(3x-1)(27-x)}=0 \\ \\ \frac{27-x-3x^2+x}{(3x-1)(27-x)}=0 \\ \\ \frac{-3x^2+27}{(3x-1)(27-x)}=0 \\ \\ \frac{3x^2-27}{(3x-1)(27-x)}=0 \\ \\ \frac{3(x^2-9)}{(3x-1)(27-x)}=0 \\ \\ \frac{3(x-3)(x+3)}{(3x-1)(27-x)}=0 \\ \\ \left \{ {{3(x-3)(x+3)=0} \atop {{(3x-1)(27-x) \neq 0}} \right. \\ \\1)x -3=0;x_1=3 \\ 2)x+3=0;x_2=-3$