因式分解&展開&恆等式

2010-09-04 5:10 pm
因式分解下列各式(列步驟)

1.(2a-b)^2-(a-7b)^2


2.6a^2+3ab-8a-4b
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展開下列各式(列步驟)

1.(2a-7b)^2

2.(5+2y^3)(5-2y^3)

3.(6a-5b)^2-(a-2b)^2

4.(x+1)^2(x-1)^2
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證明以下恆等式
1.3(y+1)+2(y+6)三5(y+3)

2.(a-b)(c-d)+(a-d)(b-c)三(a-c)(b-d)

回答 (2)

2010-09-04 5:36 pm
✔ 最佳答案
1)
(2a - b)^2 - (a - 7b)^2

= (2a - b - a + 7b)(2a - b + a - 7b)

= (a + 6b)(3a - 8b)

2)
6a^2 + 3ab - 8a - 4b

= 3a(2a + b) - 4(b + 2a)

= (3a - 4)(2a + b)

1)
(2a - 7b)^2

= 4a^2 - 2(2a)(7b) + 49b^2

= 4a^2 - 28ab + 49b^2

2)
(5 + 2y^3)(5 - 2y^3)

= (5)^2 - (2y^3)^2

= 25 - 4y^6

3)
(6a - 5b)^2 - (a - 2b)^2

= (6a - 5b - a + 2b)(6a - 5b + a - 2b)

= (5a - 3b)(7a - 7b)

= 35^2 - 56ab + 21b^2

4)
(x + 1)²(x - 1)²

= [(x + 1)(x - 1)]^2

= (x^2 - 1)^2

= x^4 - 2x^2 + 1

1)
L.H.S. = 3(y + 1) + 2(y + 6)

= 3y + 2y + 3 + 12

= 5y + 15

= 5(y + 3)

= R.H.S.

2)
L.H.S. = (a - b)(c - d) + (a - d)(b - c)

= ac - bc - ad + bd + ab - bd - ac + cd

= - bc - ad + ab + cd

R.H.S. = (a - c)(b - d)

= ab - bc - ad + cd = L.H.S.
2010-09-05 1:11 am


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