nomor 13 sm 14, sertakan cara

Materi PECAHAN ALJABAR

Nomor 13
[tex] \displaystyle \frac{x^2 + 7x + 12}{x+4} = \frac{(x+3)\cancel{(x+4)}}{\cancel{x+4}} \\ = x + 3 [/tex]

Nomor 14
[tex] \displaystyle \frac{7x^2 + 41x - 6}{2x^2 + 15x + 18} = \frac{(7x - 1)(\cancel{x + 6)}}{(2x + 3)\cancel{(x + 6)}} \\ = \frac{7x-1}{2x+3} [/tex]

Aljabar

13.)
x^2 + 7x + 12/x+4

=> (x + 4) (x + 3) / x + 4 ( coret )

=> x + 3

14.

7x^2 + 41x - 6/2x^2 + 5x + 18

=> (7x - 1) (x + 6) / (2x + 3) (x + 6) (Coret)

=> 7x - 1 / 2x + 3

15.

x^2 -7x + 6 / x^2 + x - 2

=> (x - 6) (x - 1) / (x + 2) (x - 1) ( coret)

=> x - 6 / x + 2