Tolong di jawab ya.. Terserah yg mana aja.. Kalo bisa dua2nya ya.. Makasih..

[tex] \left[\begin{array}{ccc}3\end{array}\right]\\\\ \frac{3x + 6}{|x - 1|} \ \textgreater \ 4\\3x + 6 \ \textgreater \ 4|x - 1|\\(3x + 6)^2 \ \textgreater \ [4(x - 1)]^2\\(3x + 6)^2 - [4(x - 1)]^2 \ \textgreater \ 0\\(3x + 6)^2 - (4x - 4)^2 \ \textgreater \ 0\\(7x + 2)(-x + 10) \ \textgreater \ 0\\ -\frac{2}{7} \ \textless \ x \ \textless \ 10\\\\\boxed{\boxed{\,\text{x = (0,1,2,3, ... ,9) = 9 bilangan yang memenuhi}}}\\\\\text{Jawaban : E}\\\\ [/tex]

[tex] \left[\begin{array}{ccc}4\end{array}\right]\\\\\text{sin } \alpha = \frac{1}{ \sqrt{7} }\\\text{cos } \alpha = \frac{ \sqrt{6} }{ \sqrt{7} } \\\\\text{Cos}\,\,\alpha = \frac{a . b}{|a| |b|}\\ \frac{ \sqrt{6} }{ \sqrt{7} } = \frac{ \sqrt{30} }{ \sqrt{5} |b| }\\ \frac{ \sqrt{6} }{ \sqrt{7} } = \frac{ \sqrt{6} }{|b|}\\|b| = \sqrt{7}\\b . b = \sqrt{7} [/tex]

3) (3x + 6)/|x - 1| > 4
|x - 1| =
1) x - 1 jika x ≥ 1
2) -(x - 1) = 1 - x jika x < 1

1) jika x ≥ 1
(3x + 6)/(x - 1) > 4
(3x + 6)/(x - 1) - 4 > 0
(3x + 6 - 4(x - 1))/(x - 1) > 0
(3x + 6 - 4x + 4)/(x - 1) > 0
(10 - x)/(x - 1) > 0
x = 10 atau x = 1
Garis bilangan
---- (1) +++ (10) ----
1 < x < 10 => memenuhi syarat x ≥ 1
HP = 1 < x < 10
Bilangan bulat yang memenuhi {2, 3, 4, 5, 6, 7, 8, 9} = 8 bilangan bulat

2) jika x < 1
(3x + 6)/(1 - x) > 4
(3x + 6)/(1 - x) - 4 > 0
(3x + 6 - 4(1 - x))/(1 - x) > 0
(3x + 6 - 4 + 4x)/(1 - x) > 0
(7x + 2)/(1 - x) > 0
x = -2/7 atau x = 1
Garis bilangan
---- (-2/7) +++ (1) ----
-2/7 < x < 1 => memenuhi syarat x < 1
HP = -2/7 < x < 1
Bilangan bulat yang memenuhi = {0} = 1 bilangan bulat
Jadi banyak bilangan bulat x = 8 + 1 = 9 bilangan bulat

4) sin ¢ = 1/√7 = depan/miring
samping = √(√7^2 - 1^2) = √(7 - 1) = √6
cos ¢ = samping/miring = √6/√7

a.b = √30
|a| |b| cos ¢ = √30
√5 . |b| . √6/√7 = √30
√30/√7 |b| = √30
|b| = √30 . √7/√30 = √7

b . b = |b|^2 = √7^2 = 7