求解一元二次不等式1.求根号9-x^2+根号6x-x^2>3的解2.k>1时,求x^2/2-x
问题描述：
求解一元二次不等式
1.求根号9-x^2+根号6x-x^2>3的解
2.k>1时,求x^2/2-x
答
1.√(9-x^2)+√(6x-x^2)>3
(9-x^2)≥0,即-3≤x≤3;
(6x-x^2)≥0,即0≤x≤6;此时0≤x≤3.
∵√(9-x^2)+√(6x-x^2)≤2√[(√(9-x^2))^2+(√(6x-x^2))^2/2]
=√(-4x^2+12x+18)
∴√(-4x^2+12x+18)>3
即-4x^2+12x+9>0
∴(3-3√2)/2