Question

fie fractia x la 3 -3x la 2-x+3 supra 9-x la 2 pentru x apartine R

Answer 1

(x³-3x²-x+3)/(9-x²)=
[x²(x-3)-(x-3)]/[(3-x)(3+x)]=
[(x-3)(x²-1)]/[-(x-3)(-x-3)]=simplif cu x-3=
(x²-1)/[(-(-x-3)]=
(x²-1)/(x+3)
cond.x+3≠0⇒x≠-3
x²-1=0
x²=1
x₁=1
x₂=-1
x∈{-1,1}\{-3}