Question

Помогите срочно, пожалуйста. 8 класс алгебра

Answer 1

4x² - 26x + 12 = 0

2x² - 13x + 6 = 0

D = (- 13)² - 4 * 2 * 6 = 169 - 48 = 121 = 11²

$x_{1}=\frac{13+11}{4}=6\\\\x_{2}=\frac{13-11}{4}=0,5$

4x² - 26x + 12 = 4(x - 6)(x - 0,5)

x³ - 216 = x³ - 6³ = (x - 6)(x² + 6x + 36)

$\frac{4x^{2} -26x+12}{x^{3}-216 }=\frac{4(x-6)(x-0,5)}{(x-6)(x^{2}+6x+36) }=\frac{4(x-0,5)}{x^{2}+6x+36 }=\frac{4x-2}{x^{2}+6x+36 }$

Answer 2

(4х²-26х+12)/(х³-216)

4х²-26х+12=0
Д1=169-48=121=11²
х=(13±11)/4
х1=6;х2=1/2

(4х²-26х+12)/(х³-6³)=
4(х-6)(х-1/2)/(х-6)(х²+6х+36)=
(4х-2)/(х²+6х+36)