Question

упростите выражение
3/x-3 + x+15/9-x^2 -1/x

Answer 1

$\tt\displaystyle\frac{3}{x-3}+\frac{x+15}{9-x^2}-\frac{1}{x}=\\\\\\-\frac{3}{3-x} +\frac{x+15}{9-x^2}-\frac{1}{x} = \\\\\\=\frac{-3\cdot x(3+x)+(x+15)\cdot x-1\cdot(3-x)(3+x)}{x(3-x)(3+x)} =\\\\\\=\frac{-3\cdot(3x+x^2)+x^2+15x-(9-x^2)}{x(3-x)(3+x)}=\\\\\\=\frac{-9x-3x^2+x^2+15x-9+x^2}{x(3-x)(3+x)} =\\\\\\=\frac{-x^2+6x-9}{x(3-x)(3+x)}=\frac{(-1)\cdot(x^2-6x+9)}{x(3-x)(3+x)} =\\\\\\=\frac{(-1)\cdot(x-3)^2}{x(3-x)(3+x)} = \frac{(-1)\cdot(x-3)(x-3)}{x(3-x)(3+x)}=$

$\tt\displaystyle=\frac{(3-x)(x-3)}{x(3-x)(3+x)}=\frac{x-3}{x(3+x)}$