(1)|x-1|+(y+1/3)²=0
因为|x-1|≥0
(y+1/3)²≥0
所以,要使它们的和为0,则它们两者都要为0
x-1=0
x=1
y+1/3=0
y=-1/3
2x²-1/2[5y²-2(x²-2y²)+6]
=2x² - 1/2[9y²-2x²+6]
=3x² - 9y²/2 -3
=3 - 1/2 -3
=-1/2
(2) (2x²-3x²y-2xy²)-(x²-2xy²+4)+(-x²+3x²y+5)
=2x²-3x²y-2xy² - x²+2xy²-4 -x²+3x²y+5
=(2x²-x²-x²) + (-3x²y + 3x²y) +(-2xy² + 2xy²) - 4 +5
=0 +0 +0 +1
=1
所以,与x和y的值无关