Respuesta :

sqdancefan

Answer:

  [tex]\dfrac{x^2-4}{x^2-6x+9}[/tex]

Step-by-step explanation:

We assume you want to expand ...

  [tex]\dfrac{x+2}{x-3}\cdot\dfrac{x-2}{x-3}=\dfrac{(x+2)(x-2)}{(x-3)^2}=\boxed{\dfrac{x^2-4}{x^2-6x+9}}[/tex]

_____

In each case, the product of the factors is ...

  (x +a)(x +b) = x² +(a+b)x +ab

For the numerator, you have (a, b) = (2, -2).

For the denominator, you have (a, b) = (-3, -3).

Hi1315

Answer:

[tex]= \frac{ {x}^{2} - 4 }{ {x }^{2}-6x+9 } \\ [/tex]

Step-by-step explanation:

[tex] \frac{x + 2}{x - 3} \times \frac{x - 2}{x - 3} \\ \frac{(x + 2)(x - 2)}{(x - 3)(x - 3)} \\ \frac{x(x - 2) + 2(x - 2)}{ {(x - 3)}^{2} } \\ \frac{ {x}^{2} - 2x + 2x - 4 }{ {x}^{2} - 6x+9} \\ = \frac{ {x}^{2} - 4 }{ {x }^{2} -6x+9 } [/tex]

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