please利用恆等式因式分解
9a3-25ab2
p2-q2+4q+4p
maths
2006-10-14 6:22 am
回答 (3)
2006-10-14 6:43 am
✔ 最佳答案
9a3-25ab2Assume your question is 9a^3-25ab^2
= 3^2a^3-5^2ab^2
= a(3a)^2-a(5b)^2
= a [ (3a)^2-(5b)^2 ]
= a [ (3a)+(5b) ] [ (3a)-(5b) ]
p2-q2+4q+4p
Assume your question is p^2-q^2+4q+4p
= (p+q)(p-q)+4(p+q)
= (p+q)(p-q+4)
2006-10-14 7:24 pm
9a³-25ab²
=a(9a²-25b²)
=a[(3a)²-(5b)²]
=a(3a+5b)(3a-5b)
p²-q²+4q+4p
=(p+q)(p-q)+4(q+p)
=(p+q)(p-q)+4(p+q)
=(p+q)[(p-q)+4]
=(p+q)(p-q+4)
=a(9a²-25b²)
=a[(3a)²-(5b)²]
=a(3a+5b)(3a-5b)
p²-q²+4q+4p
=(p+q)(p-q)+4(q+p)
=(p+q)(p-q)+4(p+q)
=(p+q)[(p-q)+4]
=(p+q)(p-q+4)
參考: Calculation
2006-10-14 8:39 am
9a^3-25ab^2
=a (9a^2-25b^2)
=a (3a - 5b) (3a + 5b)
p^2 - q^2 + 4q + 4p
= (p + q) (p - q) + 4 (q + p)
= (p + q) (p - q) + 4 (p + q)
= (p + q) (p - q + 4)
=a (9a^2-25b^2)
=a (3a - 5b) (3a + 5b)
p^2 - q^2 + 4q + 4p
= (p + q) (p - q) + 4 (q + p)
= (p + q) (p - q) + 4 (p + q)
= (p + q) (p - q + 4)
參考: myself
收錄日期: 2021-04-12 22:50:02
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