f(x) = 3·a·x/(x + 3)
g(x) = (1 + x)/(1 - x)
3·a·x/(x + 3) > 1
3·a·x/(x + 3) - 1 > 0
(x·(3·a - 1) - 3)/(x + 3) > 0
2 possibilità:
{x·(3·a - 1) - 3 > 0
{x + 3 > 0
se:
[x > -3 ∧ IF(a > 1/3, x > 3/(3·a - 1)),x > -3 ∧ IF(a < 1/3, x < 3/(3·a - 1))]
{x·(3·a - 1) - 3 < 0
{x + 3 < 0
se:
[x < -3 ∧ IF(a > 1/3, x < 3/(3·a - 1)), x < -3 ∧ IF(a < 1/3, x > 3/(3·a - 1))]
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a = 1
f (x)= 3·x/(x + 3) ; g(x) = (x + 1)/(1 - x)
0 < 3·x/(x + 3) + 3·(x + 1)/(1 - x) ≤ 1
(5·x + 3)/((x - 1)·(x + 3)) < 0 ≤ 1
x < -3 ∨ - 3/5 < x < 1