Question

Se consideră expresia E(x) = $(\frac{ x^{2}+4x+4}{ x^{2} -x-6}+\frac{ x^{2} +4x+3}{ x^{2} -2x-3}) : \frac{(2x+5)(2x+1)}{6(x-3)}$, unde x ∈ R/{-2; $-\frac{5}{2}$; -1; $\frac{1}{2}$; 3}
a) Arătaţi ca: E(x) = $\frac{6}{2x+1}$
b) Determinaţi valorile întregi ale lui x, pentru care E(x) ∈ Z.

Answer 1

[(x²+4x+4)/(x²-x-6)+(x²+4x+3)/(x²-2x-3)]:(2x+5)(2x+1)/6(x-3)=
=[(x+2)²/(x²+2x-3x-6)+(x²+3x+x+3)/(x²+x-3x-3)]  × 6(x-3)/(2x+5)(2x+1)=
=[(x+2)²/(x+2)(x-3)+(x+3)(x+1)/(x+1)(x-3)] × 6(x-3)/(2x+5)(2x+1)=
=[(x+2)/(x-3)+(x+3)/(x-3)] × 6(x-3)/(2x+5)(2x+1)=
=(2x+5)/(x-3) × 6(x-3)/(2x+5)(2x+1)=
=6/(2x+1)


x=0
6/(2x+1)=6/1=6
x=-2
6/(2x+1)=6/)(-4+1)=-2