設u=3^x，若y=(7+9u)/(u+2)，求值域?
回答 (3)
2019-05-18 5:35 pm
設u=3^x，若y=(7+9u)/(u+2)，求值域?
Sol
y=(7+9u)/(u+2)
yu+2y=7+9u
yu+2y－7－9u=0
(y－9)u=7－2y
u=(2y－7)/(y－9)>0
(2y－7)(y－9)>0
3.5<y<9
Sol
y=(7+9u)/(u+2)
yu+2y=7+9u
yu+2y－7－9u=0
(y－9)u=7－2y
u=(2y－7)/(y－9)>0
(2y－7)(y－9)>0
3.5<y<9
2019-06-08 9:20 pm
y=(9u+18-11)/(u+2)
=9-11/(u+2)
值域為 {y∈R∣7 / 2 < y < 9} #
=9-11/(u+2)
值域為 {y∈R∣7 / 2 < y < 9} #
收錄日期: 2021-04-18 18:16:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190517164537AA0J32p