Respuesta :

gmany
[tex]\dfrac{x^2+9x+18}{x+2}\cdot\dfrac{x^2-3x-10}{x^2+2x-24}=\dfrac{x^2+6x+3x+18}{x+2}\cdot\dfrac{x^2-5x+2x-10}{x^2+6x-4x-24}\\\\=\dfrac{x(x+6)+3(x+6)}{x+2}\cdot\dfrac{x(x-5)+2(x-5)}{x(x+6)-4(x+6)}[/tex]

[tex]=\dfrac{(x+6)(x+3)}{x+2}\cdot\dfrac{(x-5)(x+2)}{(x+6)(x-4)}=\dfrac{(x+3)(x-5)}{x-4}\\\\=\dfrac{x^2-5x+3x-15}{x-4}=\dfrac{x^2-2x-15}{x-4} [/tex]
applelulu

The answer is [tex]\frac{(x+3)(x-5)}{x-4} , x\neq-6, x\neq-2,  x\neq4[/tex]

Otras preguntas