Question

Произведение корней уравнения (2x + 8)^2 (13х – 39) = 26(4x^2 — 64)(х-3)
равно​

Answer 1

х1= -4  х2=3  х3=12

Объяснение:

(2x + 8)^2 (13х – 39) = 26(4x^2 — 64)(х-3)

(4$x^{2}$+32+64)(13x-39) = (104$x^{2}$-1664)(х-3)

54$x^{3}$-156$x^{2}$+416$x^{2}$-1248x-832x-2496 = 104$x^{3}$-312$x^{2}$-1664x+4992

54$x^{3}$+260$x^{2}$-416x-2496 = 104$x^{3}$-312$x^{2}$-1664x+4992

$x^{3}$+5$x^{2}$-8x-48 = 2$x^{3}$-6$x^{2}$-32x+96

$x^{3}$+5$x^{2}$-8x-48-2$x^{3}$+6$x^{2}$+32x-96 = 0

-$x^{3}$+11$x^{2}$+24x-144 = 0

-$x^{3}$+3$x^{2}$+8$x^{2}$-24x+48x-144 = 0

-$x^{2}$(x-3)+8x(x-3)+48(x-3) = 0

-(x-3)($x^{2}$-8x-48) = 0

-(x-3)($x^{2}$-4x-12x-48) = 0

-(x-3)(x(x+4)-12(x-4)) = 0

-(x-3)(x-4)(x-12) = 0

(x-3)(x-4)(x-12) = 0

x-3 = 0

x+4 = 0

x-12 = 0

x =3

x =-4

x =2

$x_{1}$ =3

$x_{2}$ = -4

$x_{3}$ =12