Question

帮我解两道数学题

（x-1)(x-2)(x-3)(x-4）
（x+y+2）(-x+y+2)(x-y-+2)（x+y-2)
计算啊

Answer 1

（x-1)(x-2)(x-3)(x-4)
=[(x-1)(x-4)][(x-2)(x-3)]
=(x^2-5x+4)(x^2-5x+6)
=(x^2-5x)^2+10(x^2-5x)+24
=x^4-10x^3+25x^2+10x^2-50x+24
=x^4-10x^3+35x^2-50x+24

（x+y+2）(-x+y+2)(x-y-+2)（x+y-2)
=-(x+y+2)(x+y-2)(x-y-2)(x-y+2)
=-[(x+y)^2-4][(x-y)^2-4]
=-(x^2+2xy+y^2-4)(x^2-2xy+y^2-4)
=-(x^2+y^2-4+2xy)(x^2+y^2-4-2xy)
=-[(x^2+y^2-4)^2-(2xy)^2]
=-(x^4+y^4+16+2x^2y^2-8x^2-8y^2-4x^2y^2)
=-x^4-y^4+2x^y^2+8x^+8y^2-16

Answer 2

计算什么？你就拿两个算式来，是叫我们把它分开么？