Докажите тождество:1)2)

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

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

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