1、∵(x+y+z)(x-y+z),
=(x+z+y)(x+z-y),
=[(x+z)+y][(x+z)-y],
=(A+B)(A-B),
∵B=y,
∴A=x+z.
2、(x+y-z)(x-(y-z))
假设y-z=a,则原式=(x+a)(x-a)=x²-a²
(A+B)(A-B)=A²-B²
所以A=X
B=y-z
1、∵(x+y+z)(x-y+z),
=(x+z+y)(x+z-y),
=[(x+z)+y][(x+z)-y],
=(A+B)(A-B),
∵B=y,
∴A=x+z.
2、(x+y-z)(x-(y-z))
假设y-z=a,则原式=(x+a)(x-a)=x²-a²
(A+B)(A-B)=A²-B²
所以A=X
B=y-z