ОДЗ:
а) x-6>0
x>6
b) x-10>0
x>10
В итоге ОДЗ: x>10
{x-6≥0
{18(x-10)≤(x-6)²
{18(x-10)≥0
a) x-6≥0
x≥6
b) 18(x-10)≤(x-6)²
18x-180≤x²-12x+36
-x²+18x+12x-180-36≤0
-x²+30x-216≤0
x²-30x+216≥0
D=900-4*216=36
x₁=(30-6)/2=12
x₂=(30+6)/2=18
+ - +
-------- 12 -------------
18 -----------
\\\\\\\\\ \\\\\\\\\\\\
x∈(-∞; 12]U[18; +∞)
c) 18(x-10)≥0
x-10≥0
x≥10
Объединяем ОДЗ и решения трех неравенств в систему:
{x>10
{x≥6
{x∈(-∞; 12]U[18; +∞)
{x≥10
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
-------- 10
------ 12 --------------
18 ------------
/////////////////////// ////////////////////////
x∈(10; 12]U[18; +∞)
Ответ: (10; 12]U[18; +∞)