Question

помогите очень срочно​

Answer 1

$1)\frac{a^{2}+3 }{a^{3}}-\frac{3-a}{3a}=\frac{3a^{2}+9-3a^{2}+a^{3}}{3a^{3}}=\frac{a^{3}+9}{3a^{3}}\\\\2)\frac{x}{x-1}+\frac{x}{x+1}=\frac{x^{2} x+x^{2}-x }{(x-1)(x+1)}=\frac{2x^{2}}{x^{2}-1}\\\\3)\frac{x}{x-2y}-\frac{4y^{2}}{x^{2}-2xy}=\frac{x}{x-2y}-\frac{4y^{2}}{x(x-2y)} =\frac{x^{2}-4y^{2}}{x(x-2y)}=\frac{(x-2y)(x+2y)}{x(x-2y)}=\frac{x+2y}{x}$

$4)2a+b-\frac{4ab}{2a+b}=\frac{(2a+b)^{2} -4ab}{2a+b}=\frac{4a^{2}+4ab+b^{2}-4ab}{2a+b}=\frac{4a^{2}+b^{2}}{2a+b}\\\\5)\frac{3}{x+3}+\frac{3}{x^{2}-3x}+\frac{2x}{9-x^{2}}=\frac{3}{x+3}+\frac{3}{x(x-3)}-\frac{2x}{(x-3)(x+3)}=\frac{3x^{2}-9x+3x+9-2x^{2}}{x(x-3)(x+3)}=\frac{x^{2}-6x+9}{x(x-3)(x+3)} =\frac{(x-3)^{2}}{x(x-3)(x+3)}=\frac{x-3}{x^{2}+3x }$