初一的计算题（1）32×2^-4(2)(1/7)^0÷(-1/7)^-1(3)-2x^2y(3xy^2z-2y^2z)(
1个回答
(1)32×2^-4
=32/2……4=32/16
=2
(2)(1/7)^0÷(-1/7)^-1
=1÷(-7)
=-1/7
(3)-2x^2y(3xy^2z-2y^2z)
=-2x^2y*3xy^2z+2x^2y*2y^2z
=-6x^3y^3z+4x^2y^3z
(4)(9x-2y)(x+y)
=9x^2+9xy-2xy-2y^2
=9x^2+7xy-2y^2
(5)(-1/5a^3x^4-9/10a^2x^3)÷(1/2πrh)
=-1/5a^3x^4÷(1/2πrh)-9/10a^2x^3÷(1/2πrh)
=-2/5a^3x^4/πrh-9/5a^2x^3/πrh
(6)(-2x+3)(-2x-3)
=(-2x)^2-3^2
=4x^2-9
(7)(-1/2x+2y)^2
=1/4x^2-2xy+4y^2
(8)(3mn+1/2)(3mn-1/2)-m^2n^2
=9m^2n^2-1/4-m^2n^2
=8m^2n^2-2/4
(9)x^2-(x+2)(x-2)
=x^2-(x^2-4)
=x^2-x^2+4
=4
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