f（x）是定义在R上的偶函数，当x＜0时，xf′（x）-f（x）＜0，且f（-3）=0，则不等式
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设函数g(x)=
，则g′(x)=f(x) x
，xf′(x)-f(x) x2
当x＜0时，xf′（x）-f（x）＜0，所以此时g′(x)=
＜0，即函数g（x）单调递减．xf′(x)-f(x) x2
又函数g(x)=
为奇函数．f(x) x
所以函数g（x）在x＞0时单调递减，且f（3）=0．
画出函数g(x)=
的草图（只体现单调性），f(x) x
则不等式
＞0的解为0＜x＜3或x＜-3．f(x) x
即不等式的解集为（-∞，-3）∪（0，3）．
故答案为：（-∞，-3）∪（0，3）．
| 完形填空。 | |||
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