（x-1）²+ │y+1│=0 求2x²y-（2x²y-4x²y+5xy²)的值RT

（x-1）²+ │y+1│=0 求2x²y-（2x²y-4x²y+5xy²)的值RT
（x-1）²+ │y+1│=0 求2x²y-（2x²y-4x²y+5xy²)的值
RT

（x-1）²+ │y+1│=0 求2x²y-（2x²y-4x²y+5xy²)的值RT
因为（x-1)≥0│y+1│≥0
（x-1）²+ │y+1│=0
所以x=1,y=-1
2x²y-（2x²y-4x²y+5xy²) =4x2y-5xy2
=xy（4x-5y）
=-1*9
=-9

x=1,y=-1
2x²y-（2x²y-4x²y+5xy²)=-9

因（x-1)≥0，│y+1│≥0，且 （x-1）²+ │y+1│=0
故x=1，y=-1
故 原式=4x2y-5xy2=-4-5=-9

∵(x-1)^2≥0，Iy+1I≥0
又(x-1)^2+Iy+1I=0
∴x-1=0，y+1=0
解之得：x=1，y=-1
∴2x^2y-(2x^2y-4x^2y+5xy^2)=4x^2y-5xy^2
=-4-5
=-9

题意得
x-1=0 y+1=0
∴x=1 y=-1
∴原式=2x^2y-2x^2y+4x^2y-5xy^2=4x^2y-5xy^2=4*1^2*(-1)-5*1*(-1)^2=-4-5=-9