因式分(1)12a2bc-24ab2c+18abc2;(2)x2-6x+5;(3)(y+2)(y+4)+1;(4)-a4b4+8a2b2-16.
题型:解答题难度:一般来源:不详
因式分 (1)12a2bc-24ab2c+18abc2; (2)x2-6x+5; (3)(y+2)(y+4)+1; (4)-a4b4+8a2b2-16. |
答案
(1)12a2bc-24ab2c+18abc2=6abc(2a-4b+3c)
(2)x2-6x+5=(x-1)(x-5)
(3)(y+2)(y+4)+1, =y2+6y+9, =(y+3)2;
(4)-a4b4+8a2b2-16, =-(a4b4-8a2b2+16), =-(a2b2-4)2, =-(ab+2)2(ab-2)2. |
举一反三
已知x-2y=5,xy=1.求下列各式的值: (1)2x2y-4xy2 (2)-x2+4xy-4y2. |
(a2+b2)2-4a2b2分解因式的结果是______ |
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