已知二次函数f(x)的二次项系数为负数,且对于任意x,恒有且f(2-x)=f(2+x)成立.解不等式f[log1/2(x

1.对于任意x,恒有且f(2-x)=f(2+x)成立
显然该二次函数对称轴为x=2 且开口向下
又有f[log1/2(x^2+x+1/2)]|2-log1/2(2x^2-x+5/8)|
|log2(4/(x^2+x+1/2)|>|log2(4/(2x^2-x+5/8))|
[1](4/(x^2+x+1/2)>(4/(2x^2-x+5/8)>1
x^1/2+x^(-1/2)=3 …………（1）
1、（1）式两边取平方,即（x^1/2+x^(-1/2)）^2 = 3^2,化简得
x + x^(-1) + 2 = 9
即x + x^(-1) = 7 ………………（2）
2、（1）式两边取立方,即（x^1/2+x^(-1/2)）^3 = 3^3,化简得
x^3/2 + x^(-3/2) + 3x *x^(-1/2) + 3x^(1/2) *x^(-1)
= x^3/2 + x^(-3/2) + 3x^(1/2) + 3x^(-1/2)
= x^3/2 + x^(-3/2) + 3(x^(1/2) + x^(-1/2))
= x^3/2 + x^(-3/2) + 9 = 3^3 = 27
所以x^3/2 + x^(-3/2) = 18
3、（2）式两边取平方,即（x + x^(-1)）^2 = 7^2,化简得
x^2+x^(-2) + 2 = 49
即x^2+x^(-2) = 47
[x^3/2+x^(-3/2)+2]/[x+x^(-1)+3]
= (18 +2) / (7+3)
= 2
3.(a^3x+a^-3x)/(a^x+a^-x)
=(a^x+a^-x)(a^2x-1+a^-2x)(a^x+a^-x)
=a^2x-1+a^-2x
=√2+1-1+1/(√2+1)
=2√2-1