(3x²-12)/(5+x)≤0 ОДЗ: 5+х≠0 х≠-5.
3*(x²-4)/(5+x)≤0 |÷3
(x-2)(x+2)/(x+5)≤0
-∞____-____-5____+____-2____-____2____+____+∞ ⇒
Ответ: x∈(-∞;-5)U[-2;2].
X²-14x+66-3-2x<0
x²-16x+63<0
x1+x2=16 U x1*x2=63
x1=7 U x2=9
+ _ +
----------------(7)---------------------(9)--------------------
x∈(7;9)
X = arc tg(-1/√3) + πk , k ∈Z
x = - π/6 + πk , k∈Z