把下列各式分解因式:(1)x4-7x2+1;(2)x4+x2+2ax+1-a2(3)(1+y)2-2x2(1-y2)+x4(1-y)2(4)x4+2x3+3x2
题型:解答题难度:一般来源:不详
把下列各式分解因式: (1)x4-7x2+1; (2)x4+x2+2ax+1-a2 (3)(1+y)2-2x2(1-y2)+x4(1-y)2 (4)x4+2x3+3x2+2x+1. |
答案
(1)x4-7x2+1 =x4+2x2+1-9x2 =(x2+1)2-(3x)2 =(x2+3x+1)(x2-3x+1);
(2)x4+x2+2ax+1-a2 =x4+2x2+1-x2+2ax-a2 =(x2+1)-(x-a)2 =(x2+1+x-a)(x2+1-x+a);
(3)(1+y)2-2x2(1-y2)+x4(1-y)2 =(1+y)2-2x2(1-y)(1+y)+x4(1-y)2 =(1+y)2-2x2(1-y)(1+y)+[x2(1-y)]2 =[(1+y)-x2(1-y)]2 =(1+y-x2+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)2. |
举一反三
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