已知（x+y-10)²+根号xy-6=0,求x³y-2x²y²+xy³的值.
问题描述：
已知（x+y-10)²+根号xy-6=0,求x³y-2x²y²+xy³的值.
答
∵(x+y-10)²+√(xy-6)=0∴x+y-10=0,xy-6=0∴x+y=10,xy=6x³y+2x²y²+xy³=xy(x²+2xy+y²)=xy(x+y)²=6×10²=6×100=600明白请采纳,有新问题请求助,