Question

100 puncte!!
scrieti sub forma de fractie algebrica ireductibila (Y-5)^2/3Y+18 : 3Y^2-75/Y^2-36
Ajutor va roog

Answer 1

$\frac{(y-5)^2}{3y+18} : \frac{3y^2-75}{y^2-36}$

=$\frac{(y-5)^2}{3(y+6)} * \frac{(y-6)(y+6)}{3(y-5)(y+5)}$

=$\frac{(y-5)(y-6)}{9(y+5)} = \frac{y^2-11y+30}{9y+45}$