[tex]\it a)\ E = \dfrac{x}{(x-3)^2}:\dfrac{x-3+x+3-3x}{(x-3)(x+3)}=\dfrac{x}{(x-3)^2}\cdot\dfrac{(x-3)(x+3)}{-x}=\\ \\ =-\dfrac{x+3}{x-3}=\dfrac{x+3}{3-x}\\ \\ \\ b)\ E = \dfrac{a+4}{a-3}\cdot\dfrac{a+5+a^2-3a-4a-16}{(a+4)(a-3)}\cdot(a-3)^2=a^2-6a-11[/tex]
Am folosit faptul că:
[tex]\it a^2+a-12=(a+4)(a-3)[/tex]
