Răspuns:
Explicație pas cu pas:
[tex]\dfrac{x-2}{x^{2}+1}+\dfrac{x^{2}+1}{x-2}=-\dfrac{5}{2}\\Fie~t=\dfrac{x-2}{x^{2}+1},~~atunci~\dfrac{x^{2}+1}{x-2}=\dfrac{1}{t} .~Obtinem~ecuatia~~~t+\dfrac{1}{t}=-\dfrac{5}{2}~|*2t[/tex]
Obtinem, 2t²+5t+2=0, Δ=5²-4·2·2=25-16=9
t1=(-5-3)/(2·2)=-2, iar t2=(-5+3)/(2·2)=-1/2. Revenim la variabila x
[tex]\dfrac{x-2}{x^{2}+1} =-2,[/tex] ⇒x-2=-2(x²+1) ⇒ x-2=-2x²-2⇒ 2x²+x=0 ⇒ x(2x+1)=0 deci x=0 sau x=-1/2.
[tex]\dfrac{x-2}{x^{2}+1}=\dfrac{-1}{2}[/tex] ⇒2(x-2)=-1(x²+1) ⇒ 2x-4=-x²-1 ⇒ x²+2x-3=0, Δ=4+12=16
x=(-2-4)/2=-3 sau x=(-2+4)/2=1
Raspuns: S={-3; -1/2; 0; 1}
