㏒₃√₂(x-5)㏒₃√₂(x+12)≤2 ОДЗ: x-5>0x>5 x+12>0 x>-12 x∈(5;+∞)
㏒₃√₂((x-5)(x+12))≤2
x²+7x-60≤(3√2)²
x²+7x-60≤18
x²+7x-78≤0
x²+7x-78=0 D=361
x₁=-13 x₂=6
(x+13)(x-5)≤0
-∞_______+________-13_______-________6________+________+∞
x∈[-13;6]
Учитывая ОДЗ:
x∈(5;6].