题目内容
分解因式:
(1)x9+x6+x3﹣3;
(2)(m2﹣1)(n2﹣1)+4mn;
(3)(x+1)4+(x2﹣1)2+(x﹣1)4;
(4)a3b﹣ab3+a2+b2+1.
(1)x9+x6+x3﹣3;
(2)(m2﹣1)(n2﹣1)+4mn;
(3)(x+1)4+(x2﹣1)2+(x﹣1)4;
(4)a3b﹣ab3+a2+b2+1.
(1)(x﹣1)(x2+x+1)(x6+2x3+3)
(2)(mn+m﹣n+1)(mn﹣m+n+1)
(3)(3x2+1)(x2+3)
(4)(a2﹣ab+1)(b2+ab+1)
(2)(mn+m﹣n+1)(mn﹣m+n+1)
(3)(3x2+1)(x2+3)
(4)(a2﹣ab+1)(b2+ab+1)
试题分析:(1)首先将﹣3拆成﹣1﹣1﹣1,多项式变为(x9﹣1)+(x6﹣1)+(x3﹣1),然后分别利用公式法分解因式即可求解;
(2)首先将4mn拆成2mn+2mn,多项式变为(m2n2+2mn+1)﹣(m2﹣2mn+n2),然后分别利用公式法分解因式即可求解;
(3)首先将(x2﹣1)2拆成2(x2﹣1)2﹣(x2﹣1)2,多项式变为[(x+1)4+2(x+1)2(x﹣1)2+(x﹣1)4]﹣(x2﹣1)2,然后利用公式法分解因式即可求解;
(4)首先添加两项+ab﹣ab,多项式变为(a3b﹣ab3)+(a2﹣ab)+(ab+b2+1),然后分别分解因式,接着提取公因式即可求解.
解:(1)原式=x9+x6+x3﹣1﹣1﹣1
=(x9﹣1)+(x6﹣1)+(x3﹣1)
=(x3﹣1)(x6+x3+1)+(x3﹣1)(x3+1)+(x3﹣1)
=(x3﹣1)(x6+2x3+3)
=(x﹣1)(x2+x+1)(x6+2x3+3);
(2)原式=(m2﹣1)(n2﹣1)+2mn+2mn
=m2n2﹣m2﹣n2+1+2mn+2mn
=(m2n2+2mn+1)﹣(m2﹣2mn+n2)
=(mn+1)2﹣(m﹣n)2
=(mn+m﹣n+1)(mn﹣m+n+1);
(3)原式=(x+1)4+2(x2﹣1)2﹣(x2﹣1)2+(x﹣1)4
=[(x+1)4+2(x+1)2(x﹣1)2+(x﹣1)4]﹣(x2﹣1)2
=[(x+1)2+(x﹣1)2]2﹣(x2﹣1)2
=(2x2+2)2﹣(x2﹣1)2=(3x2+1)(x2+3);
(4)原式=a3b﹣ab3+a2+b2+1+ab﹣ab
=(a3b﹣ab3)+(a2﹣ab)+(ab+b2+1)
=ab(a+b)(a﹣b)+a(a﹣b)+(ab+b2+1)
=a(a﹣b)[b(a+b)+1]+(ab+b2+1)
=[a(a﹣b)+1](ab+b2+1)
=(a2﹣ab+1)(b2+ab+1).
点评:此题主要考查了利用分组分解法分解因式,其中(4)是一道较难的题目,由于分解后的因式结构较复杂,所以不易想到添加+ab﹣ab,而且添加项后分成的三项组又无公因式,而是先将前两组分解,再与第三组结合,找到公因式.这道题目使我们体会到拆项、添项法的极强技巧所在,同学们需多做练习,积累经验.
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