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计算:
(1)(x+y-z)(x+y+z).
(2)(x+y-z)(x-y-z).
(3)(x+y+z)(x-y-z).
(4)(x+y-z)(x-y+z).
人气:273 ℃ 时间:2020-06-17 11:39:19
解答
(1)原式=[(x+y)-z][(x+y)+z]
=(x+y)2-z2
=x2+2xy+y2-z2
(2)原式=[(x-z)+y][(x-z)-y]
=(x-z)2-y2
=x2-2xz+z2-y2
(3)原式=[x+(y+z)][x-(y+z)]
=x2-(y+z)2
=x2-y2-z2-2yz;
(4)原式=[x+(y-z)][x-(y-z)]
=x2-(y-z)2
=x2-y2-z2+2yz.
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