[Risolto] Equazione 2 grado

(x-3)/(x-2)+(x+2)/(x-3) = 5/(x^2-5x+6)

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$\dfrac{x-3}{x-2} + \dfrac{x+2}{x-3} = \dfrac{5}{x^2-5x+6}$ → $(x^2-5x+6) = (x-2)(x-3)$ 

$\dfrac{x-3}{x-2} + \dfrac{x+2}{x-3} = \dfrac{5}{(x-2)(x+3)}$

$mcm= (x-2)(x-3)$ $(C.E.→x\not=2; x\not=3)$

per cui:

$(x-3)(x-3)+(x+2)(x-2) = 5$

$(x-3)^2+x^2-4 = 5$

$x^2-6x+9 +x^2-4 = 5$

$2x^2-6x +5 = 5$

$2x^2-6x = 5-5$

$2x^2-6x = 0$

dividi tutto per 2:

$x^2-3x = 0$

$\quad \Downarrow$

$x(x-3) = 0$

$x_1 → x= 0$

$x_2 → x-3 = 0 → x= 3$

 

$(x_2= 3$ non accettabile v. C.E.; quindi $x= 0$