Ada yg bisa bantu? Thanks

Jika |x| > a maka x < -a atau x > a
Jika |x| < a maka -a < x < a

log (1 - |t|) terdefinisi jika
1 - |t| > 0
1 > |t|
|t| < 1
-1 < t < 1
-1 < (x^2 - 3)/(3x + 7) < 1
1) (x^2 - 3)/(3x + 7) < 1
(x^2 - 3)/(3x + 7) - 1 < 0
(x^2 - 3 - (3x + 7))/(3x + 7) < 0
(x^2 - 3x - 10)/(3x + 7) < 0
(x - 5)(x + 2)/(3x + 7) < 0
x = 5 atau x = -2 atau x = -7/3
Garis bilangan
--- (-7/3) +++ (-2) --- (5) ++++
x < -7/3 atau -2 < x < 5

2)(x^2 - 3)/(3x + 7) > -1
(x^2 - 3)/(3x + 7) + 1 > 0
(x^2 - 3 + 3x + 7)/(3x + 7) > 0
(x^2 + 3x + 4)/(3x + 7) > 0 ====> x^2 + 3x + 4 definit positif karena D < 0
+ / (3x + 7) > 0
3x + 7 > 0
3x > -7
x > -7/3

Irisan (1) dan (2) : -2 < x < 5