√（x^3-x²y+xy²/4）+√(x²y/4-xy²+y^3）
=√[x(x²-xy+y²/4)]+√[y(x²/4-xy+y²)]
=√[x(x-y/2)²]+√[y(x/2-y)²],
∵X=6,y=4,∴x-y/2=4,x/2-y=-1＜0,
原式=(x-y/2)√x+(y-x/2)√y
=(6-4/2)√6+(4-6/2)√4
=4√6+√4
=2+4√6..