Respuesta :

krismlazar

You need to factor the numerator and denominator...

(2x^2+4x-2x-4)/(2x^2-2x-2x+2)

(2x(x+2)-2(x+2))/(2x(x-1)-2(x-1))

((2x-2)(x+2))/((2x-2)(x-1))  so the (2x-2)s cancel out leaving

(x+2)/(x-1)

gmany

Answer:

[tex]\large\boxed{\dfrac{2x^2+2x-4}{2x^2-4x+2}=\dfrac{x+2}{x-1}}[/tex]

Step-by-step explanation:

[tex]\dfrac{2x^2+2x-4}{2x^2-4x+2}=\dfrac{2(x^2+x-2)}{2(x^2-2x+1)}=\dfrac{x^2+x-2}{x^2-2x+1}\\\\=\dfrac{x^2+2x-x-2}{x^2-x-x+1}=\dfrac{x(x+2)-1(x+2)}{x(x-1)-1(x-1)}\\\\=\dfrac{(x+2)(x-1)}{(x-1)(x-1)}\qquad\text{cancel}\ (x-1)\\\\=\dfrac{x+2}{x-1}[/tex]

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