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

Question

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

Answer 1

(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/...