|x+1|+(y-2)²=0|x+1|≥0(y-2)²≥0所以|x+1|=0,(y-2)²=0x=-1,y=23xy²-2(x²y-2xy)+5x²y-xy²=2xy²+3x²y+2xy=xy(2y+3x+2)=(-1)*2*[2*2+3*(-1)+2]=-6