factorize each of the followings.
-a^3 – b^6
36a^6 + 48a^3b^2 + 16b^4
81h^4 - 72h^2 + 16
f.2 maths
2007-04-30 11:53 pm
回答 (3)
2007-05-01 12:06 am
✔ 最佳答案
-a3 - b6= - (a3 + b6)
= - (a + b2) (a2 - ab2 - b4)
36a6 + 48a3b2 + 16b4
= 4 (9a6 + 12a3b2 + 4b4)
= 4 (3a3 + 2b2)2
81h4 - 72h2 + 16
= (9h2 - 4)2
= (3h + 2)2 (3h - 2)2
2007-05-01 3:05 am
-a^3 – b^6
=(-a^3+b^6)(-a^3-b^6)
36a^6 + 48a^3b^2 + 16b^4
=4[9a^6+12a^3b^2+4b^4]
=4(3a^3+2b^2)^2
81h^4 - 72h^2 + 16
=(9h^2+4)^2
2007-04-30 19:07:40 補充：
81h^4 - 72h^2 + 16 =(9h^2+4)^2 =[(3h+2)^2]^2=(3h+2)^4
=(-a^3+b^6)(-a^3-b^6)
36a^6 + 48a^3b^2 + 16b^4
=4[9a^6+12a^3b^2+4b^4]
=4(3a^3+2b^2)^2
81h^4 - 72h^2 + 16
=(9h^2+4)^2
2007-04-30 19:07:40 補充：
81h^4 - 72h^2 + 16 =(9h^2+4)^2 =[(3h+2)^2]^2=(3h+2)^4
2007-05-01 12:14 am
-a^3 – b^6
= -[a^3 + (b^2)^3]
= -(a + b^2)(a^2 - ab^2 + b^4) //
36a^6 + 48a^3b^2 + 16b^4
= 4 (9a^6 + 12a^3b^2 + 4b^4)
= 4 [(3a^3)^2 + (2)(3a^2)(2b)^2 + (2b^2)^2
= 4 (3a^3 + 2b^2)^2 //
81h^4 - 72h^2 + 16
= (9h^2)^2 - 2(9h^2)(4) + 4^2
= (9h^2 - 4)^2
= [(3h)^2 - 2^2]^2
= [(3h - 2)(3h + 2)]^2
= (3h - 2)^2 (3h + 2)^2 //
= -[a^3 + (b^2)^3]
= -(a + b^2)(a^2 - ab^2 + b^4) //
36a^6 + 48a^3b^2 + 16b^4
= 4 (9a^6 + 12a^3b^2 + 4b^4)
= 4 [(3a^3)^2 + (2)(3a^2)(2b)^2 + (2b^2)^2
= 4 (3a^3 + 2b^2)^2 //
81h^4 - 72h^2 + 16
= (9h^2)^2 - 2(9h^2)(4) + 4^2
= (9h^2 - 4)^2
= [(3h)^2 - 2^2]^2
= [(3h - 2)(3h + 2)]^2
= (3h - 2)^2 (3h + 2)^2 //
收錄日期: 2021-04-13 00:24:46
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