若实数xy满足x≥y≥2,且2x²-xy-8x+2y+9=0,则根号xy的值
问题描述：
若实数xy满足x≥y≥2,且2x²-xy-8x+2y+9=0,则根号xy的值
答
原式可变成x^2-xy+x^2-6x+9-2x+2y=x(x-y)+(x-3)^2-2(x-y)=(x-2)(x-y)+(x-3)^2=0因为x≥y≥2,所以x-2≥0,x-y≥0所以(x-2)(x-y)=0,（x-3)^2=0解得x=3,y=3根号xy=3 楼上接着算就好啦,根据题目得到x≥2(x-2)+1/(x-2)≥2...