Risolvere la seguente equazione frazionaria di primo grado:

$\displaystyle \frac{1}{x}+\frac{1}{x-2}=\frac{2}{x-3}$

$\displaystyle \frac{(x-2)(x-3)+x(x-3)}{x(x-2)(x-3)}=\frac{2[x(x-2)]}{x(x-2)(x-3)}$

$\displaystyle \frac{x^2-3x-2x+6+x^2-3x}{x(x-2)(x-3)}=\frac{2x^2-4x}{x(x-2)(x-3)}$

C.E: $x\neq 0$, $x\neq 2$, $x\neq 3$

$\displaystyle x^2-3x-2x+6+x^2-3x=2x^2-4x$

$\displaystyle -3x-2x-3x+4x=-6$

$\displaystyle -4x=-6$

$\displaystyle \frac{-4x}{-4}=\frac{-6}{-4}$

$\displaystyle x=\frac{6}{4}=\frac{3}{2}$