tolong bantu jawab soal nomor 3 ya, kak thanks before^^

a)
[tex] (x+2)^{x+4} = (x+2)^{ x^{2} +3x+1} [/tex]

=> x + 4 = x² + 3x + 1
     x² + 3x + 1 - x - 4 = 0
     x² + 2x - 3 = 0
     (x + 3) (x - 1) = 0
     x1 = -3;  x2 = 1

b). x + 2 = 1
     x3 = 1 - 2 
          = -1

Jadi himpunan penyelesaiannya adalah {-3, -1, 1}  ⇒ jawaban D

(x + 2)^(x + 4) = (x + 2)^(x^2 + 3x + 1)

1) x + 2 = 1
=> x = 1 - 2
=> x = -1

2) x + 2 = 0 jika (x + 4) > 0 dan (x^2 + 3x + 1) > 0
=> x = 0 - 2
=> x = -2
(x + 4) = (-2 + 4) = 2 > 0
(x^2 + 3x + 1) = (-2)^2 + 3(-2) + 1 = 4 - 6 + 1 = -1 => 0^-1 = 1/0^1 = 1/0 = ?
karena (x^2 + 3x + 1) < 0 maka
x = -2 tidak memenuhi

3) x + 2 = -1 jika (x + 4) dan (x^2 + 3x + 1) sama2 ganjil atau sama2 genap
=> x = -1 - 2
=> x = -3
(x + 4) = (-3 + 4) = 1 => ganjil
(x^2 + 3x + 1) = (-3)^2 + 3(-3) + 1 = 9 - 9 + 1 = 1 => ganjil
Jadi x = -3 memenuhi

4) (x + 4) = (x^2 + 3x + 1)
=> 0 = x^2 + 2x - 3
=> (x + 3)(x - 1) = 0
=> x = -3 atau x = 1

HP = {-3, -1, 1}