(完整word版)因式分解过关练习题及答案(2).pdf

因式分解专题过关1将下列各式分解因式（1）3p26pq（2）2x2+8x+8 2将下列各式分解因式（1）x3yxy（2）3a36a2b+3ab23分解因式（1）a2（xy）+16（yx）（2）（x2+y2）24x2y2 4分解因式：（1）2x2x（2）16x21（3）6xy29x2yy3（4）4+12（xy）+9（xy）2 5因式分解：（1）2am28a（2）4x3+4x2y+xy2 6将下列各式分解因式：（1）3x12x3（2）（x2+y2）24x2y2 7因式分解：（1）x2y2xy2+y3（2）（x+2y）2y2 8对下列代数式分解因式：（1）n2（m2）n（2m）（2）（x1）（x3）+1 9分解因式：a24a+4b210分解因式：a2b22a+1 11把下列各式分解因式：（1）x47x2+1（2）x4+x2+2ax+1a2（3）（1+y）22x2（1y2）+x4（1y）2（4）x4+2x3+3x2+2x+1 12把下列各式分解因式：（1）4x331x+15；（2）2a2b2+2a2c2+2b2c2a4b4c4；（3）x5+x+1；（4）x3+5x2+3x9；（5）2a4a3 6a2a+2文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4因式分解专题过关1将下列各式分解因式（1）3p26pq；（2）2x2+8x+8 分析：（1）提取公因式3p 整理即可；（2）先提取公因式2，再对余下的多项式利用完全平方公式继续分解解答：解：（1）3p26pq=3p（p2q），（2）2x2+8x+8，=2（x2+4x+4），=2（x+2）22将下列各式分解因式（1）x3yxy（2）3a36a2b+3ab2分析：（1）首先提取公因式xy，再利用平方差公式进行二次分解即可；（2）首先提取公因式3a，再利用完全平方公式进行二次分解即可解答：解：（1）原式=xy（x21）=xy（x+1）（x1）；（2）原式=3a（a22ab+b2）=3a（ab）23分解因式（1）a2（xy）+16（yx）；（2）（x2+y2）24x2y2分析：（1）先提取公因式（xy），再利用平方差公式继续分解；（2）先利用平方差公式，再利用完全平方公式继续分解解答：解：（1）a2（xy）+16（y x），=（xy）（a216），=（xy）（a+4）（a4）；（2）（x2+y2）24x2y2，=（x2+2xy+y2）（x22xy+y2），=（x+y）2（xy）24分解因式：（1）2x2x；（2）16x21；（3）6xy29x2yy3；（4）4+12（xy）+9（xy）2文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4分析：（1）直接提取公因式x 即可；（2）利用平方差公式进行因式分解；（3）先提取公因式y，再对余下的多项式利用完全平方公式继续分解；（4）把（xy）看作整体，利用完全平方公式分解因式即可解答：解：（1）2x2x=x（2x1）；（2）16x21=（4x+1）（4x1）；（3）6xy29x2y y3，=y（9x26xy+y2），=y（3xy）2；（4）4+12（xy）+9（xy）2，=2+3（xy）2，=（3x3y+2）25因式分解：（1）2am28a；（2）4x3+4x2y+xy2 分析：（1）先提公因式2a，再对余下的多项式利用平方差公式继续分解；（2）先提公因式x，再对余下的多项式利用完全平方公式继续分解解答：解：（1）2am28a=2a（m24）=2a（m+2）（m2）；（2）4x3+4x2y+xy2，=x（4x2+4xy+y2），=x（2x+y）26将下列各式分解因式：（1）3x12x3（2）（x2+y2）24x2y2分析：（1）先提公因式3x，再利用平方差公式继续分解因式；（2）先利用平方差公式分解因式，再利用完全平方公式继续分解因式解答：解：（1）3x12x3=3x（14x2）=3x（1+2x）（12x）；（2）（x2+y2）24x2y2=（x2+y2+2xy）（x2+y22xy）=（x+y）2（xy）27因式分解：（1）x2y2xy2+y3；（2）（x+2y）2y2分析：（1）先提取公因式y，再对余下的多项式利用完全平方式继续分解因式；（2）符合平方差公式的结构特点，利用平方差公式进行因式分解即可解答：解：（1）x2y2xy2+y3=y（x22xy+y2）=y（xy）2；（2）（x+2y）2 y2=（x+2y+y）（x+2yy）=（x+3y）（x+y）文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H48对下列代数式分解因式：（1）n2（m2）n（2m）；（2）（x 1）（x 3）+1分析：（1）提取公因式n（m2）即可；（2）根据多项式的乘法把（x1）（x3）展开，再利用完全平方公式进行因式分解解答：解：（1）n2（m 2）n（2m）=n2（m2）+n（m2）=n（m2）（n+1）；（2）（x1）（x3）+1=x24x+4=（x2）29分解因式：a24a+4b2分析：本题有四项，应该考虑运用分组分解法观察后可以发现，本题中有a 的二次项a2，a 的一次项 4a，常数项4，所以要考虑三一分组，先运用完全平方公式，再进一步运用平方差公式进行分解解答：解：a24a+4 b2=（a24a+4）b2=（a 2）2b2=（a2+b）（a2b）10分解因式：a2b22a+1 分析：当被分解的式子是四项时，应考虑运用分组分解法进行分解本题中有a的二次项，a 的一次项，有常数项所以要考虑a22a+1 为一组解答：解：a2b22a+1=（a22a+1）b2=（a 1）2b2=（a1+b）（a1b）11把下列各式分解因式：（1）x47x2+1；（2）x4+x2+2ax+1a2（3）（1+y）22x2（1y2）+x4（1y）2（4）x4+2x3+3x2+2x+1 分析：（1）首先把 7x2变为+2x29x2，然后多项式变为x42x2+19x2，接着利用完全平方公式和平方差公式分解因式即可求解；（2）首先把多项式变为x4+2x2+1x2+2axa2，然后利用公式法分解因式即可解；（3）首先把 2x2（1 y2）变为 2x2（1 y）（1y），然后利用完全平方公式分解因式即可求解；文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4（4）首先把多项式变为x4+x3+x2+x3+x2+x+x2+x+1，然后三个一组提取公因式，接着提取公因式即可求解解答：解：（1）x4 7x2+1=x4+2x2+1 9x2=（x2+1）2（3x）2=（x2+3x+1）（x23x+1）；（2）x4+x2+2ax+1a=x4+2x2+1x2+2axa2=（x2+1）（x a）2=（x2+1+xa）（x2+1x+a）；（3）（1+y）22x2（1y2）+x4（1y）2=（1+y）22x2（1y）（1+y）+x4（1y）2=（1+y）2 2x2（1y）（1+y）+x2（1y）2=（1+y）x2（1 y）2=（1+yx2+x2y）2（4）x4+2x3+3x2+2x+1=x4+x3+x2+x3+x2+x+x2+x+1=x2（x2+x+1）+x（x2+x+1）+x2+x+1=（x2+x+1）212把下列各式分解因式：（1）4x331x+15；（2）2a2b2+2a2c2+2b2c2a4 b4c4；（3）x5+x+1；（4）x3+5x2+3x9；（文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4文档编码:CL10F2X1M8D5 HP1E9R7A4Q1 ZT6L10A3T5H4