Question

Descompunând membrul stâng in factori, să se rezolve ecuațiile:
1) 6x^3-x^2-486x+81=0
2) x^3+3x^2-16x+48=0

Vă rog, am nevoie urgent!

Answer 1

Răspuns:

Explicație pas cu pas:

1) 6x^3-x^2-486x+81=0

6x(x^2 - 81) - (x^2 - 81) = 0

(x^2 - 81)(6x - 1) = 0

(x - 9)(x + 9)(6x -1) = 0

x1 = 9, x2 = -9, x3 = 1/6

2) x^3+3x^2-16x+48=0

x^2 (x - 3) - 16(x - 3) = 0

(x - 3)(x^2 - 16) = 0

(x - 3)(x - 4)(x + 4) = 0 cu radacinile

x1 = 3, x2 = 4 si x3 = -4.

Answer 2

$\it 1)\ \ 6x^3-x^2-486x+81=0 \Rightarrow6x(x^2-81)-(x^2-81)=0 \Rightarrow \\ \\ \\\Rightarrow (x^2-81)(6x-1)=0 \Rightarrow \begin{cases} \it x^2-81=0 \Rightarrow x^2=81 \Rightarrow x=\pm9\\ \\ \it6x-1=0 \Rightarrow 6x=1 \Rightarrow x=\dfrac{1}{6}\end{cases}\\ \\ \\ x_1=-9,\ \ x_2=\dfrac{1}{6},\ \ x_3=9\ .$

$\it 2)\ \ x^3-3x^2-16x+48=0 \Rightarrow x(x^2-16)-3(x^2-16)=0 \Rightarrow \\ \\ \Rightarrow (x^2-16)(x-3)=0 \Rightarrow \begin{cases} \it x^2-16=0 \Rightarrow x^2=16 \Rightarrow x=\pm4\\ \\\it x-3=0 \Rightarrow x=3 \end{cases}\\ \\ \\ x_1=-4,\ \ x_2=3,\ \ x_3=4$