1：若不等式x^2+2+|x^3-2x|>=ax对x属于(0,4)恒成立,则实数a的取值范围是?

解(1)x^2+2+|x^3-2x|≥ax
因为x∈(0,4)
所以两边同时除以x
所以有：x+2/x+|x^2-2|≥a
对于x+2/x+|x^2-2|在x=√2时有最小值
所以a∈(-∞,2√2]
(2)设x=cosα,y=√2sinα,0＜α＜π/2
x√(1+y^2)=cosα√[1+2(sinα)^2]
=√[(cosα)^2+2(sinαcosα)^2]
=√{(cosα)^2+(1/2)[sin(2α)]^2}
=√{(1/2)[1+cos(2α)]+(1/2)[sin(2α)]^2}
=√(1/2)√{1+cos(2α)+[sin(2α)]^2}
=√(1/2)√{2+cos(2α)-[cos(2α)]^2}
=√(1/2)√{-[cos(2α)-1/2]^2+9/4}
0＜α＜π/2
0＜2α＜π
-1＜cos(2α)＜1
-3/2＜cos(2α)-1/2＜1/2
0≤[cos(2α)-1/2]^2＜1/4或0≤[cos(2α)-1/2]^2＜9/4
0≤[cos(2α)-1/2]^2＜9/4
-9/4＜-[cos(2α)-1/2]^2≤0
0＜9/4-[cos(2α)-1/2]^2≤9/4
0＜√{9/4-[cos(2α)-1/2]^2}≤3/2
0＜x√(1+y^2)≤(3/2)√(1/2)
最大值(3/2)√(1/2)