$\frac{x-2}{x+1}-\frac{5}{1-x}=\frac{x^2+27}{x^2-1}$
$\frac{x-2}{x+1}+\frac{5}{x-1}-\frac{x^2+27}{(x-1)(x+1)}=0$
$\frac{(x-2)(x-1)+5(x+1)-(x^2+27)}{(x-1)(x+1)}=0$
$\frac{x^2-2x-x+2+5x+5-x^2-27}{(x-1)(x+1)}=0$
$\frac{2x-20}{(x-1)(x+1)}=0$
$\{\begin{array}{ll}x-1\ne 0& \\ x+1\ne 0& \\ 2x-20=0& \end{array}$
$\{\begin{array}{ll}x\ne 1& \\ x\ne -1& \\ 2x=20& \end{array}$
$\{\begin{array}{ll}x\ne 1& \\ x\ne -1& \\ x=10& \end{array}$
Ответ: x = 10

$\frac{3x+1}{3x-1}-\frac{3x-1}{3x+1}=\frac{6}{1-9x^2}$
$\frac{3x+1}{3x-1}-\frac{3x-1}{3x+1}-\frac{6}{1-9x^2}=0$
$\frac{3x+1}{3x-1}-\frac{3x-1}{3x+1}+\frac{6}{9x^2-1}=0$
$\frac{3x+1}{3x-1}-\frac{3x-1}{3x+1}+\frac{6}{(3x-1)(3x+1)}=0$
$\frac{(3x+1{)}^2-(3x-1{)}^2+6}{(3x-1)(3x+1)}=0$
$\frac{9x^2+6x+1-(9x^2-6x+1)+6}{(3x-1)(3x+1)}=0$
$\frac{9x^2+6x+1-9x^2+6x-1+6}{(3x-1)(3x+1)}=0$
$\frac{12x+6}{(3x-1)(3x+1)}=0$
$\{\begin{array}{ll}3x-1\ne 0& \\ 3x+1\ne 0& \\ 12x+6=0& \end{array}$
$\{\begin{array}{ll}3x\ne 1& \\ 3x\ne -1& \\ 12x=-6& \end{array}$
$\{\begin{array}{ll}x\ne \frac{1}{3}& \\ x\ne -\frac{1}{3}& \\ x=-\frac{1}{2}& \end{array}$
Ответ: $x=-\frac{1}{2}$

$\frac{4}{x-3}+\frac{1}{x}=\frac{5}{x-2}$
$\frac{4}{x-3}+\frac{1}{x}-\frac{5}{x-2}=0$
$\frac{4x(x-2)+(x-3)(x-2)-5x(x-3)}{x(x-3)(x-2)}=0$
$\frac{4x^2-8x+x^2-3x-2x+6-5x^2+15x}{x(x-3)(x-2)}=0$
$\frac{2x+6}{x(x-3)(x-2)}=0$
$\{\begin{array}{ll}x\ne 0& \\ x-3\ne 0& \\ x-2\ne 0& \\ 2x+6=0& \end{array}$
$\{\begin{array}{ll}x\ne 0& \\ x\ne 3& \\ x\ne 2& \\ 2x=-6& \end{array}$
$\{\begin{array}{ll}x\ne 0& \\ x\ne 3& \\ x\ne 2& \\ x=-3& \end{array}$
Ответ: x = −3

$\frac{2x^2-2x}{x^2-4}+\frac{6}{x+2}=\frac{x+2}{x-2}$
$\frac{2x^2-2x}{(x-2)(x+2)}+\frac{6}{x+2}-\frac{x+2}{x-2}=0$
$\frac{2x^2-2x+6(x-2)-(x+2{)}^2}{(x-2)(x+2)}=0$
$\frac{2x^2-2x+6x-12-(x^2+4x+4)}{(x-2)(x+2)}=0$
$\frac{2x^2+4x-12-x^2-4x-4}{(x-2)(x+2)}=0$
$\frac{x^2-16}{(x-2)(x+2)}=0$
$\{\begin{array}{ll}x-2\ne 0& \\ x+2\ne 0& \\ x^2-16=0& \end{array}$
$\{\begin{array}{ll}x\ne 2& \\ x\ne -2& \\ x^2=16& \end{array}$
$\{\begin{array}{ll}x\ne 2& \\ x\ne -2& \\ x=\pm 4& \end{array}$
Ответ: x = ±4

$\frac{7}{x^2+2x}+\frac{x+1}{x^2-2x}=\frac{x+4}{x^2-4}$
$\frac{7}{x^2+2x}+\frac{x+1}{x^2-2x}-\frac{x+4}{x^2-4}=0$
$\frac{7}{x(x+2)}+\frac{x+1}{x(x-2)}-\frac{x+4}{(x-2)(x+2)}=0$
$\frac{7(x-2)+(x+1)(x+2)-x(x+4)}{x(x-2)(x+2)}=0$
$\frac{7x-14+x^2+x+2x+2-x^2-4x}{x(x-2)(x+2)}=0$
$\frac{6x-12}{x(x-2)(x+2)}=0$
$\frac{6(x-2)}{x(x-2)(x+2)}=0$
$\frac{6}{x(x-2)(x+2)}=0$
6 ≠ 0
Ответ: нет корней

$\frac{x^2-9x+50}{x^2-5x}=\frac{x+1}{x-5}+\frac{x-5}{x}$
$\frac{x^2-9x+50}{x(x-5)}-\frac{x+1}{x-5}-\frac{x-5}{x}=0$
$\frac{x^2-9x+50-x(x+1)-(x-5{)}^2}{x(x-5)}=0$
$\frac{x^2-9x+50-x^2-x-(x^2-10x+25)}{x(x-5)}=0$
$\frac{x^2-9x+50-x^2-x-x^2+10x-25}{x(x-5)}=0$
$\frac{-x^2+25}{x(x-5)}=0$
$\{\begin{array}{ll}x\ne 0& \\ x-5\ne 0& \\ -x^2+25=0& \end{array}$
$\{\begin{array}{ll}x\ne 0& \\ x\ne 5& \\ x^2=25& \end{array}$
$\{\begin{array}{ll}x\ne 0& \\ x\ne 5& \\ x=\pm 5& \end{array}$
Ответ: x = −5