Question

整式乘法-积的乘方习题4道

如图.

Answer 1

(-3y)^2*x^3y+(-6xy)^2*(-xy)
=9y^2*x^3y+36x^2y^2*(-xy)
=27x^3y^3-36x^3y^3
=-9x^3y^3

-6a^2b*ab^2c/2*(-2bc^2)^3
=-3a^2b*ab^2c*(-8b^bc^6)
=24a^2b*ab^2c*b^3c^6
=24a^(2+1)b^(1+2+3)c^(1+6)
=24a^3b^6c^7

[-3a^(n+2)b]^3*[-4ab^(n+3)]^2
=-27a^3(n+2)b^3*16a^2b^2(n+3)
=-432a^(3n+6)b^3*a^2b^(2n+6)
=-432a^(3n+6+2)b^(2n+6+3)
=-432a^(3n+8)b^(2n+9)

(3a^2b^2)^2*(-2ab)^3+4ab^3*18a^6b^4
=9a^4b^4*(-8a^3b^3)+72a^(1+6)b^(3+4)
=-72a^(4+3)b^(4+3)+72a^(1+6)b^(3+4)
=-72a^7b^7+72a^7b^7
=0

Answer 2

解答：
(1)(-3y)^2*x^3y+(-6xy)^2*(-xy)=-27x^3y^3-36x^3y^3=-9x^3y^3
(2)-6a^2b*ab^2c/2*(-2bc^2)^3=24a^(2+1)*b^(1+2+3)c^(1+6)=24a^3b^6c^7
(3)(-3a^(n+2)b)^3*(-4ab^(n+3))^2=-27a^(3n+6)b^3*16a^2b^(2n+6)=432a^(3n+8)b^(2n+9)
(4)(3a^2b^2)^2*(-2ab)^3+4ab^3*18a^6b^4=-72a^7b^7+72a^7b^7=0