Question

Посогите с 1 пожалуйста!!!!!

Answer 1

$1)\frac{14x^{4}y }{49x^{3}y^{2}} =\frac{2x}{7y}\\\\2)\frac{5x}{x^{2}+3x }=\frac{5x}{x(x+3)}=\frac{5}{x+3}\\\\3)\frac{x^{2}-y^{2}}{2x-2y}=\frac{(x-y)(x+y)}{2(x-y)} =\frac{x+y}{2}\\\\4)\frac{x^{2}-16 }{x^{2}-8x+16}=\frac{(x-4)(x+4)}{(x-4)^{2}}=\frac{x+4}{x-4}$