Respuesta :

PsychrolutesMarcidus
[tex] \frac{1}{x-1}+ \frac{2}{x+2}= \frac{3}{2} [/tex]
[tex]\frac{1(x+2)}{(x-1)(x+2)}+ \frac{2(x-1)}{(x+2)(x-1)}= \frac{3(x-1)(x+2)}{2}[/tex]
[tex]\frac{x+2+2(x-1)}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}[/tex]
[tex]\frac{x+2+2x-1}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}[/tex]
[tex]\frac{3x+1}{(x-1)(x+2)}= \frac{3(x^2+x-2)}{2}[/tex]
[tex]\frac{}{(x-1)(x+2)}= \frac{3x^2+3x-6)}{2}[/tex]
[tex]\frac{3x+1}{(x^2+x-2)}= \frac{3x^2+3x-6)}{2}[/tex]
[tex]\frac{2(3x+1)}{2(x^2+x-2)}= \frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}[/tex]
[tex]\frac{2(3x+1)}{2(x^2+x-2)}-\frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}=0[/tex]
[tex]\frac{2(3x+1)-3x^2+3x-6}{2(x^2+x-2)}=0[/tex]

this will be continued

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