当(x+1)²+(y+1)²=0时,求3(2x²-xy-y²)-2(x²-2xy-3y²)的值
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当(x+1)²+(y+1)²=0时,求3(2x²-xy-y²)-2(x²-2xy-3y²)的值
当(x+1)²+(y+1)²=0时,求3(2x²-xy-y²)-2(x²-2xy-3y²)的值
当(x+1)²+(y+1)²=0时,求3(2x²-xy-y²)-2(x²-2xy-3y²)的值
x=-1,y=-1;所求式子等于8
这很简单,两个同次幂的单项式,之和为零,肯定他们都是零,所以X=Y=-1,所以后面的式子=3×[2×﹙-1﹚-1-1]-2×[-1-2+3]
8