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
绝对值和开平方所得的都是非负数，所以两项相加等于0必定两项都等于0，即：
x+y-7=0
xy-6=0
从而x=1,y=6;或x=6,y=1.代入得：
(x+2y)/(y-x)=2.6或(x+2y)/(y-x)=-1.6
答
|x+y-7|+√(xy-6) =0 => x+y-7=0; xy=6 => (7-y)y=6 => y^2-7y+6=0=(y-1)(y-6)
=> y=1;x=6 or y=6,x=1
=>（x+2y)/(y-x)=-8/5 or 13/5...ans
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