Log(x-3)(x²-12x+36)≤0
x-3>0⇒x>3
x-3≠1⇒x≠4
(x-6)²>0⇒x<6 U x>6
x∈(3;4) U (4;6) U (6;∞)
1)x∈(3;4)
x²-12x+36≥1
x²-12x+35≥0
x1+x2=12 U x1*x2=35⇒x1=5 U x2=7
x≤5 U x≥7
x∈(3;4)
2)x∈(4;6) U (6;∞)
x²-12x+36≤1
x²-12x+35≤0
x1+x2=12 U x1*x2=35⇒x1=5 U x2=7
5≤х≤7
x∈[5;6) U (6;7]
Ответ x∈(3;4) U [5;6) U (6;7]