Question

$\frac{x^{2}-x }{4x +12}$ ÷ $\frac{15 x^{2} }{14}$

Answer 1

$\frac{ \frac{ x^{2} -x}{4x +12} }{ \frac{15 x^{2} }{14} } = \frac{ \frac{x(x-1)}{4(x+3)} }{ \frac{15 x^{2} }{14} } = \frac{14x(x-1)}{4.15 x^{2} (x+3)} = \frac{7(x-1)}{30x(x+3)}$

Listo no se puede hacer mas.