Question

求解一道初二数学题(急!!!!!)

已知双曲线y=5/x与直线y=x-7有一交点为(a,b)，求a/b+b/a的值

Answer 1

http://www.pkuschool.com/zadmin/manage/details.asp?TopicAbb=directions&FileName=c2e3sxb433aa02.htm
例题7

Answer 2

交点在双曲线和直线上，所以有
b=5/a b=a-7
变形一下
ab=5 a-b=7
所以
a/b+b/a
=(a*a+b*b)/(ab)
=[(a-b)(a-b)+2ab]/(ab)
=(7*7+2*5)/5
=59/5

Answer 3

a/b+b/a=[(a-b)(a-b)+2ab]/ab
因为y=5/x
所以xy=5 ab=5
y=x-7 x-y=7 (a-b)(a-b)+2ab=(x-y)(X-y)=2*5=49+10=59
结果是59/5

Answer 4

11.8