Question

Ofer coroana.Urgeententa

Answer 1

c)
$\frac{x+1}{x^2-x}-\frac{x+2}{2x^2-2}+\frac{1}{2x-2x^3} = \frac{x+1}{x(x-1)} - \frac{x+2}{2(x^2-1)}+\frac{1}{2x(1-x^2)} = \\ = \frac{2(x+1)^2-x(x+2)-1}{2x(x-1)(x+1)} = \frac{2(x^2+2x+1)-x^2-2x-1}{2x(x-1)(x+1)} = \frac{2(x+1)^2-(x+1)^2}{2x(x-1)(x+1)} = \\ \frac{(x+1)^2}{2x(x-1)(x+1)} = \frac{x+1}{2x^2-2x}$

d)
$\frac{3}{2x-8}+\frac{3-2x}{2x^2+8x}-\frac{15x-12}{x^3-16x} = \frac{3}{2(x-4)}+\frac{3-2x}{2x(x+4)}-\frac{15x-12}{x(x^2-16)} = \\ = \frac{3x(x+4)+(3-2x)(x-4)-2(15x-12)}{2x(x-4)(x+4)} = \frac{3x^2+12x+(3x-12-2x^2+8x)-30x+24}{2x(x-4)(x+4)} = \\ =\frac{3x^2+12x+3x-12-2x^2+8x-30x+24}{2x(x-4)(x+4)}=\frac{x^2-7x+12}{2x(x-4)(x+4)} = \frac{x^2-4x-3x+12}{2x(x-4)(x+4)}= \\ = \frac{x(x-4)-3(x-4)}{2x(x-4)(x+4)}=\frac{(x-4)(x-3)}{2x(x-4)(x+4)}=\frac{x-3}{2x(x+4)}= \frac{x-3}{2x^2+8x}$