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(1)(-3x^2y)^3*xyz*(-1/3xy)^2 (2)(x-2y+4)(x+2y-4) (3)(-3)^0+(-1/2)^-2/l-2l
人气:183 ℃ 时间:2020-05-07 01:25:06
解答
(1)(-3x^2y)^3*xyz*(-1/3xy)^2=(-27x^6y^3)×xyz×1/9x^2y^2=-3x^9y^6z(2)(x-2y+4)(x+2y-4) =[x-(2y-4)][x+(2y-4)]=x²-(2y-4)²=x²-4y²+16y-16 (3)(-3)^0+(-1/2)^-2/l-2l=1+4/2=3
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