設u=3^x，若y=(7+9u)/(u+2)，求值域?

Question

螞蟻雄兵 · Answer

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

? · Answer

y=(9u+18-11)/(u+2)
=9-11/(u+2)
值域為 {y∈R∣7 / 2 < y < 9} #

照家 · Answer

https://scontent.ftpe8-1.fna.fbcdn.net/v/t1.15752-9/60513157_400059457390503_5747960354348990464_n.png?_nc_cat=108&_nc_ht=scontent.ftpe8-1.fna&oh=a5e8d3f645a975b0820f1942a11f758a&oe=5D6F5CB8

設u=3^x，若y=(7+9u)/(u+2)，求值域?

3 個解答