[tex]\displaystyle 1)\frac{4}{a}+\frac{a-4}{a+3}=\frac{4(a+3)+a(a-4)}{a(a+3)}=\frac{4a+12+a^2-4a}{a^2+3a}=\frac{12+a^2}{a^2+3a}\\ \\2)\frac{2x^2}{x^2-4}-\frac{2x}{x+2}=\frac{2x^2}{(x-2)(x+2)}-\frac{2x}{x+2}=\frac{2x^2-2x(x-2)}{(x-2)(x+2)}=\frac{2x^2-2x^2+4x}{x^2-4}=\\\frac{4x}{x^2-4}[/tex]