How to factorize b^2+6b-9-a^2
回答 (4)
2015-08-04 6:55 am
b² + 6b - 9 - a²
= b² + 2(3)b - 3² - a²
= (b - 3)² - a²
= (b - 3 + a)(b - 3 - a)
= (a + b - 3)(-a + b - 3)
= -(a + b - 3)(a - b + 3)
2015-08-03 23:39:10 補充：
嘻嘻～
做錯了～
請指教～
2015-08-03 23:43:18 補充：
似係題目打錯了～
若題目改了：
b² + 6b + 9 - a²
= b² + 2(3)b + 3² - a²
= (b + 3)² - a²
= (b + 3 + a)(b + 3 - a)
= (a + b + 3)(-a + b + 3)
= -(a + b + 3)(a - b - 3)
= b² + 2(3)b - 3² - a²
= (b - 3)² - a²
= (b - 3 + a)(b - 3 - a)
= (a + b - 3)(-a + b - 3)
= -(a + b - 3)(a - b + 3)
2015-08-03 23:39:10 補充：
嘻嘻～
做錯了～
請指教～
2015-08-03 23:43:18 補充：
似係題目打錯了～
若題目改了：
b² + 6b + 9 - a²
= b² + 2(3)b + 3² - a²
= (b + 3)² - a²
= (b + 3 + a)(b + 3 - a)
= (a + b + 3)(-a + b + 3)
= -(a + b + 3)(a - b - 3)
2015-08-04 7:48 am
題目應該打錯了 無法factoriae的
2015-08-04 7:25 am
(b - 3)² = b² - 6b + 9
b² + 6b - 9 ≠ (b - 3)²
b² + 6b - 9 ≠ (b - 3)²
2015-08-04 10:54 pm
By completing square method:
b^2+6b-9-a^2
=[b^2+(3)(2b)+9-9]-9-a^2
=(b+3)^2-18-a^2
=(b+3)^2-(a^2+18)
=(b+3)^2-(a^2+[(sqrt18)/2]a-[(sqrt18)/2]a+18)
=(b+3)^2-(a+(sqrt18))^2+[(sqrt18)/2]a
=(b+3+a+[(sqrt18)/2])(b+3-a-[(sqrt18)/2])+[(sqrt18)/2]a
END
b^2+6b-9-a^2
=[b^2+(3)(2b)+9-9]-9-a^2
=(b+3)^2-18-a^2
=(b+3)^2-(a^2+18)
=(b+3)^2-(a^2+[(sqrt18)/2]a-[(sqrt18)/2]a+18)
=(b+3)^2-(a+(sqrt18))^2+[(sqrt18)/2]a
=(b+3+a+[(sqrt18)/2])(b+3-a-[(sqrt18)/2])+[(sqrt18)/2]a
END
收錄日期: 2021-04-29 21:57:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150803000051KK00077