Trebuie sa afla cum se descompun cele doua ecuatii(numaratorul si numitorul).
O ecuatie de forma ''ax²+bx+c''se descompune a(x-x₁)(x-x₂),unde x₁ si x₂
sunt solutiile ecuatiei...
14x²+x-3=0
Δ=1+14*4*3
Δ=169⇒√Δ=13
x₁=-1-13/28=-1/2
x₂=-1+13/28=3/7
Deci ecuatia va fi descompusa:14(x+1/2)(x-3/7)
Luam a doua ecuatie si facem la fel:
7x²+18x-9=0
Δ=324+4*9*7=324+252=576⇒√Δ=24
x₁=-18+24/14 = 6/14= 3/7
x₂=-18-24/14 = -42/14 = -3\\
Aceasta se va descompune:7(x-3/7)(x+3)
Revenind:
E(x)=14(x+1/2)(x-3/7) / 7(x-3/7)(x+3)
E(x)=2(x+1/2) / x+3
E(x)=2x+1/ x+3
x+3 / x+3 ⇒x+3/ 2x+6
x+3 /2x+1⇒x+3/ 2x+1
⇒x+3 / 5⇒ x+3 ∈D5
x+3 ∈{-5,-1,1,5}
x∈ {-8,-4,-2,2}
