问题
解答题
(1)求不等式的解集:-x2+4x+5<0 (2)求函数的定义域:y=
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答案
(1)∵-x2+4x+5<0,∴x2-4x-5>0,∴(x-5)(x+1)>0,解得x<-1或x>5,即解集为{x|x<-1或x>5};
(2)令
≥0,则x-1 x+2
,解得x<-2或x≥1,即定义域为{x|x<-2或x≥1}.(x-1)(x+2)≥0 x+2≠0
(1)求不等式的解集:-x2+4x+5<0 (2)求函数的定义域:y=
|
(1)∵-x2+4x+5<0,∴x2-4x-5>0,∴(x-5)(x+1)>0,解得x<-1或x>5,即解集为{x|x<-1或x>5};
(2)令
≥0,则x-1 x+2
,解得x<-2或x≥1,即定义域为{x|x<-2或x≥1}.(x-1)(x+2)≥0 x+2≠0