ОДЗ 6x-x²≥0⇒x(6-x)≥0
x∈[0;6]
√(6x-x²)≥0⇒1/(x-3)(x-4)-(x-4)/(x-3)≤0
(1-x²+8x-16)/9x-3)(x-4)≤0
(x²-8x+15)/(x-3)(x-4)≥0
x²-8x+15=0
x1+x2=8 U x1*x2=15⇒x1=3 u x2=5
(x-3)(x-5)/(x-3)(x-4)≥0
(x-5)/(x-4)≥0,x≠3
+ _ +
------------ (4)---------[5]---------------
x<4 U x≥5<br>x∈[0;3) U (3;4) U [5;6]