题目内容
设S=(x-1)4+4(x-1)3+6(x-1)2+4(x-1)+1,则S等于( )
分析:观察S=(x-1)4+4(x-1)3+6(x-1)2+4(x-1)+1发现,均含有x-1所以令t=x-1,则S=t4+4t3+6t2+4t+1,再观察答案各项均是一个完整的四次式,因而要因式分解.分解后再将t=x-1代入分解后的因式,即可知S
解答:解:令t=x-1,
则S=t4+4t3+6t2+4t+1,
=t4+t3+3t3+3t2+3t2+3t+t+1,
=t3(t+1)+3t2(t+1)+3t(t+1)+(t+1),
=(t+1)(t3+3t2+3t+1),
=(t+1)(t3+t2+2t2+2t+t+1),
=(t+1)[t2(t+1)+2t(t+1)+(t+1)],
=(t+1)2(t2+2t+1),
=(t+1)4
再将t=x-1代入S=(t+1)4=x4.
故选C.
则S=t4+4t3+6t2+4t+1,
=t4+t3+3t3+3t2+3t2+3t+t+1,
=t3(t+1)+3t2(t+1)+3t(t+1)+(t+1),
=(t+1)(t3+3t2+3t+1),
=(t+1)(t3+t2+2t2+2t+t+1),
=(t+1)[t2(t+1)+2t(t+1)+(t+1)],
=(t+1)2(t2+2t+1),
=(t+1)4
再将t=x-1代入S=(t+1)4=x4.
故选C.
点评:本题巧妙利用换元法,拆分项,提取公因式法,本题虽是选择题,仍可做为大题出现.同学们通过本题可以综合锻炼自己的思维与能力.
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