tolong dibantu ya fliss

1) √(2x + 6) < √(4 - x) ====> syarat : 1) 2x + 6 ≥ 0 => 2x ≥ -6 => x ≥ -3
(√(2x + 6))^2 < (√(4 - x))^2 ................ 2) 4 - x ≥ 0 => -x ≥ -4 => x ≤ 4
2x + 6 < 4 - x
2x + x < 4 - 6
3x < -2
x < -2/3
Garis bilangan
-------------- (-2/3) +++++++ => arsir daerah yg negatif
---- (-3) +++++++++++++++ => arsir daerah yg positif
-------------------------- (4) ++ => arsir daerah yg negatif
Irisan HP = -3 ≤ x < -2/3

2) √(4 - x^2) < x + 2 ===> syarat : 4 - x^2 ≥ 0
(√(4 - x^2))^2 < (x + 2)^2 .............. (2 + x)(2 - x) ≥ 0
4 - x^2 < x^2 + 4x + 4 ................... x = -2 atau x = 2
-2x^2 - 4x < 0
-2x (x + 2) < 0
x = 0 atau x = -2
Garis bilangan
-------- (-2) +++++ (0) --------------- => arsir daerah yg negatif
-------- (-2) ++++++++++ (2) ------ => arsir daerah yg positif
Irisan HP = 0 < x ≤ 2