已知(x+1)²+|Y-½|=0,求2[XY²+X²Y)-[2XY²-3﹙1-X²Y﹚]-2的值.
∵(x+1)²+|Y-½|=0
∴x=-1 y=½
2[XY²+X²Y)-[2XY²-3﹙1-X²Y﹚]-2
=2xy²+2x²y-2xy²+3-3x²y-2
=-x²y+1
=-(-1)²×½+1
=-½+1
=½
已知(x+1)²+|Y-½|=0,求2[XY²+X²Y)-[2XY²-3﹙1-X²Y﹚]-2的值.
∵(x+1)²+|Y-½|=0
∴x=-1 y=½
2[XY²+X²Y)-[2XY²-3﹙1-X²Y﹚]-2
=2xy²+2x²y-2xy²+3-3x²y-2
=-x²y+1
=-(-1)²×½+1
=-½+1
=½