1 Factorize u^2+4uv+4v^2-9.
2 Using the result of (1), factorize (u^2+4uv+4v^2)^2-81.
F.3 Factorization
2007-09-29 5:18 am
回答 (4)
2007-09-29 5:28 am
✔ 最佳答案
1. u^2 + 4uv + 4v^2 - 9
= (u^2 + 4uv + 4v^2) - 9
= [u^2 +2 x u x (2v) + (2v)^2] - 9
= (u + 2v)^2 - 9, [p.s. a^2 + 2ab + b^2 = (a + b)^2]
= (u + 2v)^2 - 3^2
= (u + 2v + 3)(u + 2v - 3), [p.s. a^2 - b^2 = (a + b)(a -b)]
2.
(u^2 + 4uv + 4v^2)^2 - 81
= (u^2 + 4uv + 4v^2)^2 - 9^2
= [(u^2 + 4uv + 4v^2) + 9] x [(u^2 + 4uv + 4v^2) - 9]
by the result of (1),
= (u^2 + 4uv + 4v^2 + 9) x [(u + 2v)^2 - 9]
= (u^2 + 4uv + 4v^2 + 9)(u + 2v + 3)(u + 2v - 3)
2007-09-29 5:32 am
1. u^2 + 4uv + 4v^2 - 9
= u^2 + 2u(2v) + (2v)^2 - 3^2
= (u + 2v)^2 - 3^2
= (u + 2v - 3) (u + 2v + 3)
2. (u^2 + 4uv + 4v^2)^2 - 81
= [(u^2 + 4uv + 4v^2) - 9] [(u^2 + 4uv + 4v^2) + 9]
= (u + 2v - 3) (u + 2v + 3) (u^2 + 4uv + 4v^2 + 9) (using the result of 1)
= u^2 + 2u(2v) + (2v)^2 - 3^2
= (u + 2v)^2 - 3^2
= (u + 2v - 3) (u + 2v + 3)
2. (u^2 + 4uv + 4v^2)^2 - 81
= [(u^2 + 4uv + 4v^2) - 9] [(u^2 + 4uv + 4v^2) + 9]
= (u + 2v - 3) (u + 2v + 3) (u^2 + 4uv + 4v^2 + 9) (using the result of 1)
2007-09-29 5:29 am
好easy
1. u^2+4uv+4v^2-9.
=(u)^2+2(u)(2v)+(2v)2-(3)^2
=(u+2v)^2-(3)^2
=(u+2v-3)(u+2v+3) .
2.(u^2+4uv+4v^2)^2-81
=(u+2v)^4-(3)^4
=((u+2v)^2+9)((u+2v)^2-9)
=((u+2v)^2+9)(u+2v+3)(u+2v-3)
1. u^2+4uv+4v^2-9.
=(u)^2+2(u)(2v)+(2v)2-(3)^2
=(u+2v)^2-(3)^2
=(u+2v-3)(u+2v+3) .
2.(u^2+4uv+4v^2)^2-81
=(u+2v)^4-(3)^4
=((u+2v)^2+9)((u+2v)^2-9)
=((u+2v)^2+9)(u+2v+3)(u+2v-3)
2007-09-29 5:23 am
1)
u^2+4uv+4v^2-9
=(u+2v)^2-(3)^2
=(u+2v-3)(u+2v+3)
2)
(u^2+4uv+4v^2)^2-81
=(u+2v)^2-(9)^2
=(u+2v-9)(u+2v+9)
u^2+4uv+4v^2-9
=(u+2v)^2-(3)^2
=(u+2v-3)(u+2v+3)
2)
(u^2+4uv+4v^2)^2-81
=(u+2v)^2-(9)^2
=(u+2v-9)(u+2v+9)
參考: me!
收錄日期: 2021-04-13 13:39:53
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