解不等式,確定取值的范圍.
2x²+ax+2＞0
=>2x²+ax+2=2(x+a/4)^2+2-a^2/8>=2-a^2/8>0 =>a^2 - 4ax^2>=2x-1 =>
a>=(2x-1)/x^2=2/x-1/x^2=-(1/x)^2+2*1/x -1 +1=-(1/x-1)^2+1
-(1/x-1)^2+1a>=2
x²-2x+1-a²≥0
=>a^2=0 => a^2 a=0
kx² -（2k+1）x - 3=0
在（-1,1）和（1,3）上有根.求K的范圍.
x=0 ,kx² -（2k+1）x - 3=-3