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

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  • 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