题目内容
(1) a3—10a2+25a (2)把9991分解成两个素数(质数)的积
(1) a(a-5)2(2)103×97
(1) a3—10a2+25a (2)把9991分解成两个素数(质数)的积
=a(a2-10a+25) 9991=10000-9
=a(a-5)2 =1002-32
=(100+3)(100-3)(2分)
=103×97
=a(a2-10a+25) 9991=10000-9
=a(a-5)2 =1002-32
=(100+3)(100-3)(2分)
=103×97
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