1.find the values of A and B in the identiy.
Ay^2-12y-B≡3(y+1)(y-5)
Expand the following expressions.
2.(4+x)(4-x)
3.(8a+3b)(8a-3b)
4.(3x+4)^2
5.(5-2a)^2
6.(4x^2+9)(2x-3)(2x+3)
7.(a)Expand (x-3k)^2
(b)Using the result of (a),expand (a+2b-3k)^2
F.2 Maths Identity(快)
2010-09-01 11:49 pm
回答 (2)
2010-09-02 3:15 am
✔ 最佳答案
1)L.H.S. = Ay^2 - 12y - B
R.H.S. = 3y^2 - 12y - 15
So, A = 3, B = 15
2)
(4 + x)(4 - x)
= 4^2 - x^2
= 16 - x^2
3)
(8a + 3b)(8a - 3b)
= (2a)^2 - (3b)^2
= 4a^2 - 9b^2
4)
(3x + 4)^2
= 9x^2 + 24x + 16
5)
(5 - 2a)^2
= 25 - 2(5)(2a) + (2a)^2
= 25 - 20a + 4a^2
6)
(4x^2 + 9)(2x - 3)(2x + 3)
= (4x^2 + 9)[(2x)^2 - 9]
= (4x^2 + 9)(4x^2 - 9)
= (4x^2)^2 - 9^2
= 16x^4 - 81
7)
(x - 3k)^2
= x^2 - 2(x)(3k) + (3k)^2
= x^2 - 6kx + 9k^2
b)
(a + 2b - 3k)^2
= (a + 2b)^2 - 6k(a + 2b) + 9k^2
= a^2 + 4ab + 4b^2 - 6ka - 12kb + 9k^2
= a^2 - (6k - 4b)a + (9k^2 - 12kb + 4b^2)
參考: Hope can help u
2010-09-02 1:02 am
1. LHS=Ay^2-12y-B
RHS=3(y+1)(y-5)
=3y^2-12y-15
A=3
B=15
2.(4+x)(4-x)
=4^2-x^2
=16-x^2
2010-09-01 17:05:18 補充：
3. (8a+3b)(8a-3b)
=(8a)^2-(3b)^2
=64a^2-9b^2
4.(5-2a)^2
=5^2-20a+(2a)^2
=25-20a+4a^2
2010-09-01 17:09:15 補充：
5.(4x^2+9)(2x-3)(2x+3)
=(4x^2+9)((2x)^2+3^2)
=(4x^2+9)(4x^2+9)
=16x^4+36x^2+81
6.(x-3k)^2
=x^2-6kx+9k^2
RHS=3(y+1)(y-5)
=3y^2-12y-15
A=3
B=15
2.(4+x)(4-x)
=4^2-x^2
=16-x^2
2010-09-01 17:05:18 補充：
3. (8a+3b)(8a-3b)
=(8a)^2-(3b)^2
=64a^2-9b^2
4.(5-2a)^2
=5^2-20a+(2a)^2
=25-20a+4a^2
2010-09-01 17:09:15 補充：
5.(4x^2+9)(2x-3)(2x+3)
=(4x^2+9)((2x)^2+3^2)
=(4x^2+9)(4x^2+9)
=16x^4+36x^2+81
6.(x-3k)^2
=x^2-6kx+9k^2
收錄日期: 2021-04-15 15:24:03
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https://hk.answers.yahoo.com/question/index?qid=20100901000051KK00801