Question

Ajuda aí galera,por favor!!!!!
Resolva as equações:
|x   4  -2|
|x-1 x  1| =  |x 3|
|1 x+1 3|    | 2 1|




|x  0  1|
|2x x 2| =0
|3 2x x|

Answer 1

Por Matriz e Determinantes:

$\left[\begin{array}{ccc}x&4&-2\\x-1&x&1\\1&x+1&3\end{array}\right] = \left[\begin{array}{ccc}x&3\\2&1\end{array}\right]$

$\left[\begin{array}{ccc}x&4&-2\\x-1&x&1\\1&x+1&3\end{array}\right | \left\begin{array}{ccc}x&4\\x-1&x\\1&x+1\end{array}\right] = \left[\begin{array}{ccc}x&3\\2&1\end{array}\right\left|\begin{array}{ccc}x\\2\end{array}\right]$


[(x . x . 3) + (4 . 1 . 1) + (-2.(x-1) . 1)] - [(-2 . x . 1) + (x . 1.(x + 1)) + (4.(x - 1) . 3)] = (x .1) - (3 . 2)

[(3x²) + (4) + ((-2x + 2) . 1)] - [(-2x + 2) . 1)  + (x² + x) + ((4x - 4) . 3)] = (x) - (6)

[3x² + 4 + (-2x + 2)] - [(-2x +2) + x² + x + (12x - 12)] = x -6
[3x² + 4 - 2x + 2] - [-2x + 2 + x² + x + 12x - 12] = x - 6
[3x² - 2x + 4 + 2] - [x² + - 2x + x + 12x + 2 - 10] = x - 6
3x² - 2x + 6 - x² - 11x + 10 = x - 6

3x² - x² - 2x - 11x + 6 + 10 = x - 6
2x² - 13x + 16 - x + 6 = 0
2x² - 13x - x + 16 + 6 =
2x² - 14x + 22 = 0  (dividindo tudo por 2)

x² - 7x + 11 = 0

a = 1
b = - 7
c = 11

Δ = b² - 4.a.c
Δ = (-7)² - 4.(1).(11)
Δ = 49 - 44
Δ = 5

x = $\frac{-b \ +- \ \sqrt{Δ} }{2a}$
x = $\frac{-(-7) \ + - \ \sqrt{5} }{2 \ . \ 1}$

x' = $\frac{7 \ + \ \sqrt{5} }{2}$


x'' = $\frac{7 \ - \ \sqrt{5} }{2}$

S = {$\frac{7 \ + \ \sqrt{5} }{2}$; $\frac{7 \ - \ \sqrt{5} }{2}$}


$\left[\begin{array}{ccc}x&0&1\\2x&x&2\\3&2x&x\end{array}\right]$

$\left[\begin{array}{ccc}x&0&1\\2x&x&2\\3&2x&x\end{array}\right | \left\begin{array}{ccc}x&0\\2x&x\\3&2x\end{array}\right] = 0$

[(x . x . x) + (0 . 2 . 3) + (1 . 2x . x)] - [(1 . x . 3) + (x . 2 . 2x) + (0 . 2x . x)] = 0
[x³ + 0 + 2x²] - [3x + 4x² + 0] = 0
[x³ + 2x²] - [3x + 4x²] = 0
x³ + 2x² - 3x + 4x² =
x³ + 2x² + 4x² - 3x = 0
x³ + 6x² - 3x = 0  (colocando o "x"  em evidência):


x . (x² + 6x - 3) = 0    (I)

x = $\frac{0}{ x^{2} \ + \ 6x \ - \ 3}$

x = 0



x² + 6x - 3 = $\frac{0}{x}$    (II)

x² + 6x - 3 = 0


Δ = (6)² - 4 . (1) . (-3)
Δ = 36 +12
Δ = 48


$x = \frac{-6 \ +- \ \sqrt{48} }{2 . 1} \\ x = \frac{-6 \ +- \ \sqrt{2^{2}.2^{2}.3} }{2} \\ x = \frac{-6 \ +- \ 2 . 2 . \sqrt{3} }{2} \\ x = \frac{-6 \ +- \ 4 \sqrt{3} }{2}$


$x' = \frac{-6 \ + \ 4 \sqrt{3} }{2} \\ x' = \frac{-6}{2} \ + \ \frac{4 \sqrt{3}}{2} \\ x' = -3 \ + \ 2 \sqrt{3}$


$x'' = \frac{-6 \ - \ 4 \sqrt{3} }{2} \\ x'' = \frac{-6}{2} \ - \ \frac{4 \sqrt{3}}{2} \\ x'' = -3 \ - \ 2 \sqrt{3}$

S = {-3 + 2√3; -3 - 2√3)