1.)Factorize the following:
a.)999^2+999
b.)101^2-101
Factorize....Math
2009-11-02 12:19 am
回答 (5)
2009-11-03 4:21 am
✔ 最佳答案
a)999^2+999=999x(999+1)
=999x1000
=999000
b)101^2-101
=101x(101-1)
=101x100
=10100
參考: me, please no copycat
2009-11-02 12:28 am
a.)999^2+999
= 999(999 + 1)
= 999 * 1000
= 999000
= 2^3 * 3^3 * 5^3 * 37
b)101^2 - 101
= 101(101 - 1)
= 101 * 100
= 10100
= 2^2 * 5^2 * 101
2009-11-01 16:33:43 補充：
題目是因式分解,並非計算。
= 999(999 + 1)
= 999 * 1000
= 999000
= 2^3 * 3^3 * 5^3 * 37
b)101^2 - 101
= 101(101 - 1)
= 101 * 100
= 10100
= 2^2 * 5^2 * 101
2009-11-01 16:33:43 補充：
題目是因式分解,並非計算。
2009-11-02 12:27 am
a) 999*(999+1)
=999000
b) 101*(101-1)
=10100
=999000
b) 101*(101-1)
=10100
2009-11-02 12:26 am
1.)Factorize the following:
a.)999^2+999
=999*999+999
=999*(999+1)
=999*1000
b.)101^2-101
=101*101-101
=101*(101-1)
=101*100
a.)999^2+999
=999*999+999
=999*(999+1)
=999*1000
b.)101^2-101
=101*101-101
=101*(101-1)
=101*100
2009-11-02 12:25 am
a.)999^2+999
b.)101^2-101
b.)101^2-101
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