Seudónimo Seudónimo
  • 17-03-2020
  • Mathematics
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please help with this q

please help with this q class=

Respuesta :

hanifchoirunnisa
hanifchoirunnisa hanifchoirunnisa
  • 17-03-2020

[tex] \frac{2 {x}^{2} - 2x }{x - 1} \div \frac{2x + 8}{x - 3} \\ = \frac{2x {}^{2} - 2x}{x - 1} \times \frac{x - 3}{2x + 8} \\ = \frac{(2 {x}^{2} - 2x)(x - 3) }{(x - 1)(2x + 8)} [/tex]

So:

1) x - 1 ≠ 0

x ≠ 1

2) 2x + 8 ≠ 0

2x ≠ -8

x ≠ -4

(B)

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