*已知y=y1-y2,y1与x成反比例,y2与〔x-2〕成正比例,且当x=3时,y=5.当x=1时,y= -1,求y与x

1、设y=ax-b(x-2)=（a-b)x+2b,则有
5=3(a-b)+2b
-1=a-b+2b
解dea=1,b=-2
y=3x-4
2、设AC与BD交予O点,由等腰梯形的对称性可知,OA=OD=√2/2AD=3√2/2
OB=OC=√2/2BC=7√2/2
S=(OA*OD+OA*OB+OC*OD+OC*OB)/2
=13