Question

求数学的学霸，过程

Answer 1

解：原式
=[3xy-(2x²-1/2y²)*[3xy+(2x²-1/2y)²]
=(3xy)²-(2x²-1/2y²)²
=9x²y²-4x^4+2x²y²-1/4y^4
=11x²y²-4x^4-1/4y^4
=11*(1/2)²*2²-4*(1/2)^4-1/4*2^4
=11-1/4-4
=6又3/4

Answer 2

x=1/y，所以xy=1 x^2=1/y^2
原式=[3/xy-(2/x^2-1/2y^2)][3/xy+(2/x^2-1/2y^2)]=(3/xy)^2-(2/x^2-1/2y^2)^2
=9-(3/2y^2)^2=9-36=-27

Answer 3

=[3xy-(2x^2-1/2y^2)][3xy+(2x^2-1/2y^2)]
=(3xy)^2-(2x^2-1/2y^2)^2
以下自己算哈。。。

Answer 4

这种题。。。代进去一算就出来了啊，就最简单的加减乘除！

Answer 5

(3xy-2x^2+1/2y^2)(3xy+2x^2-1/2y^2)
=[3xy+(-2x^2+1/2y^2)][3xy-(-2x^2+1/2y^2)]
=(3xy)^2-(-2x^2+1/2y^2)^2
=9x^2y^2-(4x^4+1/4y^4-2*2*1/2x^2y^2)
=9x^2y^2-4x^4-1/4y^4+2x^2y^2
=11x^2y^2-4x^4-1/4y^4
=11*(1/2)^2*2^2-4*(1/2)^4-1/4*2^4
=11-1/4-4
=6.75

Answer 6

为什么不直接把x=1/2,y=2代入算呢？