Question

Докажите тождество:
1)
$frac{ {x}^{3} }{ {x}^{2} - 4 } - frac{2}{x + 2} - frac{x}{x - 2} - x = - 1$
2)
$frac{1}{x(x + 1)} + frac{1}{(x + 1)(x + 2)} + frac{1}{(x + 2)(x + 3)} = frac{3}{x(x + 3)}$

Answer 1

1)

$frac{x^{3}}{x^{2}-4}-frac{2}{x+2}-frac{x}{x-2}-x=-1\frac{x^{3}}{(x-2)(x+2)}-frac{2}{x+2}-frac{x}{x-2}-x=-1\frac{x^{3}-2(x-2)-x(x+2)-x(x-2)(x+2)}{(x-2)(x+2)}=-1\frac{x^{3}-2x+4-x^{2}-2x-x(x^{2}-4)}{x^{2}-4}=-1\frac{x^{3}-2x+4-x^{2}-2x-x^{3}+4x}{x^{2}-4}=-1\frac{4-x^{2}}{x^{2}-4}=-1\frac{-(x^{2}-4)}{x^{2}-4}=-1\-1=-1$

2)

$frac{1}{x(x+1)}+frac{1}{(x+1)(x+2)}+frac{1}{(x+2)(x+3)}=frac{3}{x(x+3)}\frac{(x+2)(x+3)+x(x+3)+x(x+1)}{x(x+1)(x+2)(x+3)}=frac{3}{x(x+3)}\frac{x^{2}+3x+2x+6+x^{2}+3x+x^{2}+x}{x(x+1)(x+2)(x+3)}=frac{3}{x(x+3)}\frac{3x^{2}+9x+6}{x(x+1)(x+2)(x+3)}=frac{3}{x(x+3)}\frac{3(x^{2}+3x+2)}{x(x+1)(x+2)(x+3)}=frac{3}{x(x+3)}\frac{3(x^{2}+2x+x+2)}{x(x+1)(x+2)(x+3)}=frac{3}{x(x+3)}\frac{3(x(x+2)+(x+2))}{x(x+1)(x+2)(x+3)}=frac{3}{x(x+3)}$

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