问题
解答题
已知:x,y为实数,且y<
|
答案
依题意,得x-1≥0 1-x≥0
∴x-1=0,解得:x=1
∴y<3
∴y-3<0,y-4<0
∴|y-3|-y2-8y+16
=3-y-(y-4)2
=3-y+(y-4)
=-1.
已知:x,y为实数,且y<
|
依题意,得x-1≥0 1-x≥0
∴x-1=0,解得:x=1
∴y<3
∴y-3<0,y-4<0
∴|y-3|-y2-8y+16
=3-y-(y-4)2
=3-y+(y-4)
=-1.