Răspuns:
Explicație pas cu pas:
[tex]\dfrac{2}{x^{2}+3x}+ \dfrac{2}{x^{2}-3x}=\dfrac{2}{x(x+3)}+ \dfrac{2}{x(x-3)}=\dfrac{2(x-3)+3(x+3)}{x(x+3)(x-3)}=\dfrac{2x-6+3x+9}{x(x+3)(x-3)}=\dfrac{5x+3}{x(x+3)(x-3)}\\deci~\dfrac{5x+3}{x(x+3)(x-3)}=\dfrac{2x^{(x}}{(x+3)(x-3)}=\dfrac{2x^{2}}{x(x+3)(x-3)}\\[/tex]Atunci, ⇒5x+3=2x², de unde 2x²-5x-3=0, ec de gradul 2
Δ=(-5)²-4·2·(-3)=25+24=49>0, √Δ=7
[tex]x_{1}=\dfrac{5-7}{2*2}=\dfrac{-2}{4}=-\dfrac{1}{2}.~x_{2}=\dfrac{5+7}{4}=3\\[/tex]
Dar DVA (Domeniul Valorilor Admisibile) este R\{0,±3}.
Atunci a doua solutie nu e valabila si deci x=-1/2.
