Question

已知z²=x²+y²,化简(x+y+z)(x-y+z)(-x+y+z)(x+y-z)

Answer 1

(x+y+z)(x-y+z)(-x+y+z)(x+y-z)
={[(x+y)+z][(x+y)-z]}×{[z+(x-y)][z-(x-y)]}
=[(x+y)^2-z^2]×[z^2-(x-y)^2]
=[(x^2+y^2+2xy)-z^2]×[z^2-(x^2+y^2-2xy)]
因为z²=x²+y²
所以 将z²=x²+y²带入上式子的结果
得
(x+y+z)(x-y+z)(-x+y+z)(x+y-z)
=2xy×2xy
=4x²y²

Answer 2

(x+y+z)(x-y+z)(-x+y+z)(x+y-z)
=[(x+y+z)(x+y-z)][(x-y+z)(-x+y+z)]
=[(x+y)²-z²][z²-(x-y)²]
=(x²+2xy+y²-x²-y²)(x²+y²-x²+2xy-y²)
=(2xy)(2xy)
=4x²y²

Answer 3

(x+y+z)(x-y+z)(-x+y+z)(x+y-z)
=(x+y+z)(z+y-z)(x-y+z)(-x+y+z)
=[(x+y)²-z²][z²-(x-y)²]
=(x²+y²+2xy-z²)[z²-(x²+y²)+2xy]
=2xy*2xy
=4x²y²

Answer 4

(x+y+z)(x-y+z)(-x+y+z)(x+y-z)
=-[(x+y)²-z²][(x-y)²-z²]
=-(2xy)(-2xy)
=4x²y²

Answer 5

(x+y+z)(x-y+z)(-x+y+z)(x+y-z)
=(x+y+z)(z+y-z)(x-y+z)(-x+y+z)
=[(x+y)²-z²][z²-(x-y)²]
=[x²+y²+2xy-(x²+y²)][(x²+y²)-(x²+y²)+2xy]
=2xy*2xy
=4x²y²