已知函数f(x)满足f(x^2-1)=loga[(x^2-2)/x^2]判断奇偶性,在（1,正无穷）单调性,解f(x)=
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已知函数f(x)满足f(x^2-1)=loga[(x^2-2)/x^2]判断奇偶性,在（1,正无穷）单调性,解f(x)=loga^(1/x)

解析：设t=x^2-1==>x=±√(t+1)

F(t)=log(a,(t-1)/(t+1)) ,其定义域为(-∞,-1)(1,+∞)

F(-t)=log(a,(t+1)/(t-1))=- F(t)

∴f(x)奇函数

F(x)=log(a,(x-1)/(x+1))==> F’(x)=[(x+1)/(x-1)lna]

∵x∈(1,+∞),∴F’(x)>0

∴在区间(1,+∞)上,函数f(x)单调增；

f(x)= log(a,(x-1)/(x+1))=log(a,(1/x))

(x-1)/(x+1)=1/x

解得x1=1-√2,x2=1+√2

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