Question

初二数学分解因式

（x+y+z）2-（x-y-z)2

Answer 1

[(x+y+z)^2-(-x+y+z)^2]
={[(x+y+z)+(-x+y+z)]*[(x+y+z)-(-x+y+z)]}
=[(2y+2z)*2x]
=4xy+4xz
=4x(y+z)

Answer 2

4x(y+z)

Answer 3

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

Answer 4

解：原式=[(x+y+z)+(x-y-z)][(x+y+z)-(x-y-z)]
=(x+y+z+x-y-z)(x+y+z-x+y+z)
=2x×(2y+2z)
=4x(y+z)

Answer 5

=[（x+y+z）+(x-y-z)]*[(x+y+z)-(x-y-z)]
=2x*2(y+z)
=4xyz

Answer 6

原式=2x+2y+2z-2x+2y+2z
=4y+4z
=4(y+z)
第二种揭发：原式=2(x+y+z-x+y+z)
=2(2y+2z)
=4(y+z)