Question

Aratati ca E(x) =  [tex] \frac{6}{2x+1}

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)}
[/tex]

Answer 1

x²+4x+4=0
Δ=16-16=0
x1=x2=-4/2=-2
x²+4x+4=(x+2)²

x²-x-6=0
Δ=1+24=25
x1=(1-5)/2=-4/2=-2
x2=(1+5)/2=6/2=3
x²-x-6=(x+2)(x-3)

x²+4x+3=0
Δ=16-12=4
x1=(-4+2)/2=-2/2=-1
x2=(-4-2)/2=-3
x²+4x+3=(x+1)(x+3)

x²-2x-3=0
Δ=4+12=16
x1=(2-4)/2=-1
x2=(2+4)/2=3
x²-2x-3=(x+1(x-3)

E(x)=[(x+2)²/(x+2)(x-3)]+[(x+1)(x+3)/(x+1)(x-3)]·6(x-3)/(2x+5)(2x+1)  simplificam I fractie cu x+2 si a ii fractie cu x+1
E(x)=[(x+2)/(x-3)]+[(x+3)/(x-3)]·6(x-3)/(2x+5)(2x+1) 
E(x)=[(x+2+x+3)/(x-3)]·6(x-3)/(2x+5)(2x+1) simplificam cu x-3
E(x)=(2x+5)/1 ·1/(2x+5)(2x+1)  simplificam cu 2x+5
E(x)=1/(2x+1)
cond.2x+1≠0  ⇒x≠-1/2