1.计算题
(1)－x²·﹙－x﹚²=-x^4
(2)﹙x－2﹚﹙x＋3﹚ = x²+x-6
(3)﹙a－2b＋3c﹚﹙a＋2b－3c﹚
=a² -(2b-3c)²
=a²-4b²+12bc-9c²
﹙4﹚﹙a＋1/2﹚²·﹙a²＋1/4﹚²·﹙a－1/2﹚²
=﹙a²-1/4﹚²·﹙a²＋1/4﹚²
=﹙a^4－1/16﹚²
=a^8-1/8a^4-1/256
2.若x²＋2x＋y²－6y＋10＝0,
则（x+1)²+(y-3)²=0
x+1=0 x=-1
y-3=0 y=3