Respuesta :

frika
1.
 [tex](x+5)+ \dfrac{-2}{x-1} = \dfrac{(x+5)(x-1)-2}{x-1} = \dfrac{x^2-x+5x-5-2}{x-1} = \frac{x^2+4x-7}{x-1} [/tex] (1st part corresponedes to A)

2.
 [tex](x-1)+ \dfrac{6}{x-1} = \dfrac{(x-1)(x-1)+6}{x-1} = \dfrac{x^2-2x+1+6}{x-1} = \dfrac{x^2-2x+7}{x-1} [/tex] (2nd part correspondes to B)

3. 
[tex](2x+1)+ \dfrac{-6}{x-1} = \dfrac{(2x+1)(x-1)-6}{x-1}= \\ \dfrac{2x^2-2x+x-1-6}{x-1} = \frac{2x^2-x-7}{x-1} [/tex] (3rd part corresponds to D)


4. 
[tex] (2x-1)+\dfrac{6}{x-1}= \dfrac{(2x-1)(x-1)+6}{x-1} = \\ \dfrac{2x^2-2x-x+1+6}{x-1} = \frac{2x^2-3x+7}{x-1} [/tex] (4th part correspondes to C).

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