Question

已知x+y+z=0求（x²-y²-z²)²-4y²z²

快

Answer 1

因为x+y+z=0
所以x+y=-z x+z=-y

(x²-y²-z²)²-4y²z²
=[(x+y)(x-y)-z²]²-4y²z²
=[-z(x-y)-z²]²-4y²z²
=[-z(x-y+z)]²-4y²z²
=[-z(-y-y)]²-4y²z²
=(2zy)²-4y²z²
=4y²z²-4y²z²
=0

Answer 2

（x²-y²-z²)²-4y²z²
=（x²-y²-z²)²-（2yz）²
=（x²-y²-z²-2yz）（x²-y²-z²+2yz)
= （x²-(y-z)²)（x²-(y+z)²)
=(x-y+z)(x+y-z)(x-y-z)(x+y+z)
因为x+y+z=0
所以（x²-y²-z²)²-4y²z²=0

Answer 3

x+y+z=0
x+y=-z x+z=-y
(x²-y²-z²)²-4y²z²
=[(x+y)(x-y)-z²]²-4y²z²
=[-z(x-y)-z²]²-4y²z²
=[-z(x-y+z)]²-4y²z²
=[-z(-y-y)]²-4y²z²
=(2zy)²-4y²z²
=4y²z²-4y²z²
=0

Answer 4

（x²-y²-z²)²-4y²z²
=（x²-y²-z²)²-(2yz)²
=(x²-y²-z²+2yz)(x²-y²-z²-2yz)
=[x²-(y²-2zy+z²)][x²-(y²+2zy+z²)]
=[x²-(y-z)²][x²-(y+z)²]
=[(x+y-z)(x-y+z)][(x-y-z)（x+y+z)]
因为 x+y+z=0↑(上面!)，所以（x²-y²-z²)²-4y²z²=0

Answer 5

=0