Question

f（根号下x）=x分之1+2*根号下x

（1）求f（x）的表达式
（2）设函数g(x)=ax^2-x的平方分之1+f(x)刚是否存在实数a，使之为奇函数？
（3）解不等式f（x）-x>2
我要详解啊!

Answer 1

f(√x) = (1+2√x)/x = (1+2√x)/(√x)^2
f(x) = (1+2x)/x^2

g(x) = ax^2 - 1/x^2 + f(x)
= ax^2 - 1/x^2 + (1+2x)/x^2
= ax^2+2x/x^2
= ax^2+2/x
当a=0时，g(x)=2/x为奇函数

f(x)-x>2
(1+2x)/x^2-x>2
(1+2x)/x^2-x-2＞0
(1+2x-x^3-2x^2)/x^2＞0
1+2x-x^3-2x^2＞0
x^3+2x^2-2x-1＜0
x^3-1+2x^2-2x＜0
(x-1)(x^2+x+1)+2x(x-1)＜0
(x-1)(x^2+3x+1)＜0
(x-1)(x^2+3x+1)＜0
[x-(3-√5)/2](x-1)[x-(3+√5)/2] ＜0
x＜(3-√5)/2，或1＜x＜(3+√5)/2

追问

谢谢，你好强，但是你貌似题看错了，可能是我写的不清楚……
f(√x)=1/x+2*√x g(x)=ax^2-1/（x^）+f(x)

追答

f(√x)=1/x+2*√x = 1/(√x)^2+2√x
f(x) = 1/x^2 + 2x

g(x) = ax^2-1/(x^2) + f(x) = ax^2 - 1/(x^2) + 1/x^2 + 2x = ax^2 + 2x
当a=0时，g(x)=2x为奇函数

f(x)-x>2
1/x^2 + 2x ＞2
分母不为零，x≠0，两边同乘以x^2：
1 + x^3 ＞ 2x^2
x^3 - 2x^2 + 1 ＞0
(x^3 - x^2) - (x^2 - 1) ＞0
x^2(x-1)-(x+1)(x-1) ＞ 0
(x-1)(x^2-x-1) ＞0
{x-(1-√5)/2} (x-1) {x-(1+√5)/2} ＞0
(1-√5)/2 ＜x＜1，或 x＞(1+√5)/2