若a-b= -3 ， a^2+b^2= 6 ，試求:
(1) ab= ?
(2) a^3-b^3= ?
銜接高中 乘法公式?
2016-07-24 11:11 am
回答 (2)
2016-07-24 11:47 am
已知 a - b = -3 及 a² + b² = 6。
(1)
考慮
(a - b)² = (-3)²
a² - 2ab + b² = 9
a² + b² - 2ab = 9
6 - 2ab = 9
2ab = -3
ab = -3/2
(2)
a³ - b³
= (a - b)(a² + ab + b²)
= (a - b)(a² + b² + ab)
= (-3)(6 - 3/2)
= (-3)(9/2)
= -27/2
(1)
考慮
(a - b)² = (-3)²
a² - 2ab + b² = 9
a² + b² - 2ab = 9
6 - 2ab = 9
2ab = -3
ab = -3/2
(2)
a³ - b³
= (a - b)(a² + ab + b²)
= (a - b)(a² + b² + ab)
= (-3)(6 - 3/2)
= (-3)(9/2)
= -27/2
2016-07-25 6:32 am
1)
a-b=-3
(a-b)^2=9
a^2 -2ab+b^2=9
(a^2+b^2)-2ab=9
ab=(9-6)/(-2)
ab=-1.5
2)
a^3-b^3
=(a-b)(a^2+ab+b^2)
=(-3)(6-1.5)
=-13.5
a-b=-3
(a-b)^2=9
a^2 -2ab+b^2=9
(a^2+b^2)-2ab=9
ab=(9-6)/(-2)
ab=-1.5
2)
a^3-b^3
=(a-b)(a^2+ab+b^2)
=(-3)(6-1.5)
=-13.5
收錄日期: 2021-04-18 15:20:11
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