已知实数x、y满足关系式绝对值x+y-7绝对值+根号(xy-6)=0,求代数式(x+2y)/(y-x)的值
问题描述:
已知实数x、y满足关系式绝对值x+y-7绝对值+根号(xy-6)=0,求代数式(x+2y)/(y-x)的值
答
绝对值x+y-7绝对值+根号(xy-6)=0
x+y-7=0 x+y=7
xy-6=0 xy=6
解得x=1,y=6或x=6,y=1
(x+2y)/(y-x)=(1+12)/(6-1)=13/5
或 (x+2y)/(y-x)=(6+2)/(1-6)=-8/5