题目内容
已知公式(a+b+c)2=a2+b2+c2+2ab+2bc+2ac,试说明:
x4+y4+z4-2x2y2-2x2z2-2y2z2能被x+y+z整除.
答案:
解析:
解析:
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解:x4+y4+z4-2x2y2-2x2z2-2y2z2 =(x2)2+(y2)2+(-z2)2+2x2(-z2)+2y2(-z2)+2x2y2-4x2y2 =[x2+y2+(-z2)]2-(2xy)2 =(x2+y2-z2)2-(2xy)2 =(x2+2xy+y2-z2)(x2-2xy+y2-z2) =[(x+y)2-z2][(x-y)2-z2] =(x+y+z)(x+y-z)(x-y+z)(x-y-z) ∴所给原式能被x+y+z整除. 分析:要说明所给式能被x+y+z整除,只需将所给式分解因式的结果中含有因式x+y+z即可. |
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| 1 |
| R1 |
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| R2 |
A、R1=
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B、R1=
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C、R1=
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D、R1=
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