[(x+2y-3/2)(x-2y+3/2)+2y(2y-3)+(2^-²/3)] /(-3x)^-¹,其中x=-1,y=2007/2006
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[(x+2y-3/2)(x-2y+3/2)+2y(2y-3)+(2^-²/3)]/(-3x)^-¹,其中x=-1,y=2007/2006[(x+2y-3/2)(x-2y+3/2)+
[(x+2y-3/2)(x-2y+3/2)+2y(2y-3)+(2^-²/3)] /(-3x)^-¹,其中x=-1,y=2007/2006
[(x+2y-3/2)(x-2y+3/2)+2y(2y-3)+(2^-²/3)] /(-3x)^-¹,其中x=-1,y=2007/2006
[(x+2y-3/2)(x-2y+3/2)+2y(2y-3)+(2^-²/3)] /(-3x)^-¹,其中x=-1,y=2007/2006
原式={[x2-(2y-3/2)2]+4y2-6y+1/12}÷(1/3)
=(x2-4y2+6y-9/4+4y2-6y+1/12)÷(1/3)
=(x2-9/4+1/12)×3=-3.5
其实这道题跟y一点关系都没有
很简单嘛
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