0070163815 0070163815
  • 11-03-2021
  • Mathematics
contestada

I need help with this please. :)

I need help with this please class=
I need help with this please class=
I need help with this please class=

Respuesta :

sqdancefan
sqdancefan sqdancefan
  • 12-03-2021

9514 1404 393

Answer:

  1. 2x
  2. 2
  3. 128x +640

Step-by-step explanation:

1.

  [tex]\displaystyle\frac{8x^2-4x}{4x-2}=\frac{4x(2x-1)}{2(2x-1)}=\boxed{2x}[/tex]

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2.

  [tex]\displaystyle\frac{8x-4}{x+3}-\frac{12x-20}{2x+6}=\frac{8x-4}{x+3}-\frac{4(3x-5)}{2(x+3)}=\frac{8x-4-2(3x-5)}{x+3}\\\\=\frac{8x-4-6x+10}{x+3}=\frac{2x+6}{x+3}=\frac{2(x+3)}{x+3}=\boxed{2}[/tex]

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3.

  [tex]\displaystyle\frac{(2x+10)^7}{(x+5)^6}=\frac{(2(x+5))^7}{(x+5)^6}=2^7(x+5)^{7-6}=128(x+5)\\\\=\boxed{128x+640}[/tex]

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