Question

Помогите пожалуйста

Answer 1

$1)\frac{2x}{x+2}+\frac{4}{x^{2}+5x+6}-\frac{3}{x+3}=\frac{2x}{x+2}+\frac{4}{(x+2)(x+3)}-\frac{3}{x+3}=\frac{2x(x+3)+4-3(x+2)}{(x+2)(x+3)}=\frac{2x^{2}+6x+4-3x-6}{(x+2)(x+3)}=\frac{2x^{2}+3x-2}{(x+2)(x+3)}=\frac{2(x+2)(x-0,5)}{(x+2)(x+3)}=\frac{2x-1}{x+3}$$1)\frac{2x}{x+2}+\frac{4}{x^{2}+5x+6}-\frac{3}{x+3}=\frac{2x}{x+2}+\frac{4}{(x+2)(x+3)}-\frac{3}{x+3}=\frac{2x(x+3)+4-3(x+2)}{(x+2)(x+3)}=\frac{2x^{2}+6x+4-3x-6}{(x+2)(x+3)}=\frac{2x^{2}+3x-2}{(x+2)(x+3)}=\frac{2(x+2)(x-0,5)}{(x+2)(x+3)}=\frac{2x-1}{x+3} \\\\2)\frac{2x-1}{x+3}:\frac{2x-1}{3}=\frac{2x-1}{x+3}*\frac{3}{2x-1}=\frac{3}{x+3}\\\\3)\frac{3}{x+3}+\frac{x}{3+x}=\frac{3+x}{3+x}=1$