Question

Va rog, doresc ajutor. Multumesc mult

Answer 1

Răspuns:

Explicație pas cu pas:

$E(x)=[(\frac{x+3}{x-3} )^{2} +1+\frac{2x+6}{x-3} ]\cdot \frac{x-3}{4x} -1$

aducem la acelasi numitor in paranteza

$E(x)=[\frac{x^{2}+6x+9 }{(x-3)^{2} } +\frac{x^{2} -6x+9}{(x-3)^{2}} +\frac{(x-3)(2x+6)}{(x-3)^{2} } ]\cdot \frac{x-3}{4x} -1$

$E(x)=\frac{2x^{2} +18+2x^{2} -18}{(x-3)^{2} } \cdot \frac{x-3}{4x}-1$

$E(x)=\frac{4x^{2}}{(x-3)^{2} } \cdot \frac{x-3}{4x}-1=\frac{x}{x-3} -1=\frac{3}{x-3}$

daca x=3 expresia nu are sens pentru ca x-3=3-3=0

$\frac{3}{x-3}$∈Z⇒ x-3={1,-1,3,-3}⇒ x={4,2,6,0}