Question

Решение упражнений на сложение и вычитание алгебраических дробей.

Answer 1

$1)\frac{1}{x(x-2)} -\frac{x+1}{x^{2}-4 }=\frac{1}{x(x-2)} -\frac{x+1}{(x-2)(x+2)}=\frac{x+2-x(x+1)}{x(x-2)(x+2)}=\\\\=\frac{x+2-x^{2}-x }{x(x-2)(x+2)}=\boxed{\frac{2-x^{2} }{x(x^{2}-4) }} \\\\\\2)\frac{x-y}{4x^{2} -y^{2} }-\frac{2}{3y-6x} =\frac{x-y}{(2x-y)(2x+y)}-\frac{2}{3(y-2x)} =\frac{x-y}{(2x-y)(2x+y)}+\frac{2}{3(2x-y)}=\\\\=\frac{3*(x-y)+2*(2x+y)}{3(2x-y)(2x+y)}=\frac{3x-3y+4x+2y}{3(2x-y)(2x+y)}= \frac{7x-y}{3(2x-y)(2x+y)}=\boxed{\frac{7x-y}{3(4x^{2}-y^{2})}}$

$3)-\frac{x+2}{4-2x}-\frac{2x^{2} +4x}{x^{3} -8} =-\frac{x+2}{2(2-x)}-\frac{2x^{2} +4x}{(x -2)(x^{2}+2x+4) }=\frac{x+2}{2(x-2)}-\frac{2x^{2} +4x}{(x -2)(x^{2}+2x+4) }=\\\\=\frac{(x+2)*(x^{2}+2x+4)-2*(2x^{2}+4x)}{2(x-2)(x^{2}+2x+4) }=\frac{x^{3}+2x^{2}+4x+2x^{2}+4x+8-4x^{2}-8x}{2(x-2)(x^{2}+2x+4) }=\\\\=\boxed{\frac{x^{3}+8 }{2(x^{3}-8) }} \\\\\\4)\frac{c}{c^{2}+3c+9 }-\frac{1}{c-3} +\frac{27}{c^{3}-27 }=\frac{c}{c^{2}+3c+9 }-\frac{1}{c-3} +\frac{27}{(c-3)(c^{2}+3c+9}=$

$=\frac{c*(c-3)-(c^{2}+3c+9)+27 }{(c-3)(c^{2}+3c+9) }=\frac{c^{2}-3c-c^{2}-3c-9+27}{(c-3)(c^{2}+3c+9) }=\frac{-6c+18}{(c-3)(c^{2} +3c+9)}=\\\\=\frac{-6(c-3)}{(c-3)(c^{2}+3c+9) }=\boxed{ -\frac{6}{c^{2}+3c+9 }}$