1/(x+4)+1/(x+5)>1/(x+6)+1/(x+3)

来源：百度知道编辑：UC知道时间：2024/06/16 18:55:26

求x的取值范围
谢谢

(2x+9)/(x^2+9x+20)>(2x+9)/(x^2+9x+18)
分母，x^2+9x+20>x^2+9x+18>0恒成立。
所以，必有2x+9<0
x<-9/2.

首先分母不为0，
x≠-3，-4，-5，-6

1/(x+4)+1/(x+5)>1/(x+6)+1/(x+3)
移项
1/(x+4)-1/(x+3)>1/(x+6)-1/(x+5)
=>
[x+4-x-3]/[(x+4)(x+3)]>[x+6-x-5]/[(x+6)(x+5)]
=>
1/[(x+3)(x+4)]>1/[(x+5)(x+6)]

=>
1/(x^2+7x+12)>1/(x^2+11x+30)

=> x>-3

X不等于-4、-5、-6、-3、要不分式没意义.

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