Question

利用乘除法运算公式计算5(x+y)(x-y)-2(x+y)^2+3(x-y)^2其中x=2,y=1

Answer 1

5(x+y)(x-y)-2(x+y)^2+3(x-y)^2
=3(x-y)^2+5(x+y)(x-y)-2(x+y)^2
=(3(x-y)-(x+y))(3(x-y)+2(x+y))
=(2x-4y)(5x-y)
其中x=2,y=1
上式=(2*2-4*1)(5*2-1)
=0

Answer 2

解：原式=3(x-y)^2+5(x+y)(x-y)-2(x+y)^2
=3(x-y)^2-5(-x-y)(x-y)+2(-x-y)^2
=[(x-y)-(-x-y)][3(x-y)-2(-x-y)]
=2x(5x-y)
=10x^2-2xy

Answer 3

5(x+y)(x-y)-2(x+y)^2+3(x-y)^2
=2(x+y)(x-y-x-y)+3(x-y)(x+y+x-y)
=-4y(x+y)+6x(x-y)
=-4×1×3+6×2×1
=0