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Решите задачу:

$\frac{x-4}{x^3-x}:\Big (\frac{x-1}{2x^2+3x+1}-\frac{1}{x^2-1}\Big )=\\\\=\frac{x-4}{x(x^2-1)}:\Big (\frac{x-1}{2(x+1)(x+\frac{1}{2})}-\frac{1}{(x-1)(x+1)}\Big )=\\\\=\frac{x-4}{x(x-1)(x+1)}:\frac{(x-1)^2-2(x+\frac{1}{2})}{2(x+1)(x+\frac{1}{2})(x-1)}=\\\\=\frac{x-4}{x(x-1)(x+1)}\cdot \frac{2(x+1)(x+\frac{1}{2})(x-1)}{x^2-4x}=\\\\=\frac{x-4}{x(x-1)(x+1)}\cdot \frac{(x-1)(x+1)(2x+1)}{x(x-4)}=\frac{2x+1}{x^2}\\\\\\\star \; \; 2x^2+3x+1=0\; ,\; \; D=1\; ,\; \; x_{1,2}=\frac{-3\pm 1}{4}\; ,\\\\x_1=-1,\; x_2=-\frac{1}{2}\\\\2x^2+3x+1=2(x+1)(x+\frac{1}{2})$