<title>当n=1时,有(a-b)(a+b)=a2-b2;当n=2时,有(a-b)(a2+ab+b2)=a3-b3;当n=3时,有(a-b)(a3+a2b+ab2+b3)=a4-b4;当n=4时,有(a-b)(a4+a3b+a2b2+ab3+b4)=a5-b5;当n∈N*时,可归纳出的结论是. -知识-经验</title>
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答案【当n=1时,有(a-b)(a+b)=a2-b2;当n=2时,有(a-b)(a2+ab+b2)=a3-b3;当n=3时,有(a-b)(a3+a2b+ab2+b3)=a4-b4;当n=4时,有(a-b)(a4+a3b+a2b2+ab3+b4)=a5-b5;当n∈N*时,可归纳出的结论是.】摘要: 当n=1时,有(a-b)(a+b)=a2-b2;当n=2时,有(a-b)(a2+ab+b2)=a3-b3;当n=3时,有(a-b)(a3+a2b+ab2+b3)=a4-b4;当n=4时,有(a-b)(a4+a3b+a2b2+ab3+b4)=a5-b5;当n∈N*时,可归纳出的结论是______.
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<pubDate>10/31/2020 13:26:25</pubDate>