[數學] Factorization(急)

2010-10-12 4:00 am
我有幾條因式分解既數想問問大家,希望大家幫下幫我,唔該!THX~

1)-2px-qx+2py+qy+3ry-3rx

2a)Expand (x+b)^2
2b)Using the result of (a), expand(u^2+u+1)(u^2-u+1)
2c)Hence, find the value off 111X91 without using a calculator

3)Find the value of A,B,C in the identity
3(x^2-1)+Ax≡ (B-x)(Cx-2)

請大家列清楚D步驟同答案,唔該晒大家
更新1:

Sorry, 我有度打錯左,應該係, 2a)Expand (x+b)^2 2b)Using the result of(a), expand (x+y+z)^2

回答 (1)

2010-10-12 4:43 am
✔ 最佳答案
你好~~~

以下都是供參考而已,如有錯誤請多多指教~~

1) -2px-qx+2py+qy+3ry-3rx

= -2px - qx - 3rx + 2py + qy + 3ry

= -x(2p + q + 3r) + y(2p + q + 3r)

= (2p + q + 3r)(y - x)

2a)Expand (x+b)^2

(x + b)^2

= x^2 + 2bx + b^2

2b)Using the result of (a), expand(u^2+u+1)(u^2-u+1)

Sol: Let (u^2 + 1) be x.

(u^2 + u + 1)(u^2 - u + 1)

= (u^2 + 1 + u)(u^2 + 1 - u)

= (x + u)(x - u)

= x^2 - u^2

= (u^2 + 1)^2 - u^2

= u^4 + 2u^2 + 1 - u^2

= u^4 + u^2 + 1

2c)Hence, find the value off 111X91 without using a calculator

111 x 91

= (100 + 11) x (100 - 9)

= (100 + 1 + 10) x (100 + 1 - 10)

= (10^2 + 1 + 10)(10^2 + 1 - 10)

= (10^2 + 1)^2 - (10)^2

= 10^4 + 2(10^2) + 1^2 - 10^2

= 10^4 + 1(10^2) + 1^2

= 10000 + 100 + 1

= 10101

3)Find the value of A,B,C in the identity

3(x^2-1)+Ax≡ (B-x)(Cx-2)

Sol:

L.H.S. = 3(x^2 - 1) + Ax

= 3x^2 - 3 + Ax

= 3x^2 + Ax - 3

R.H.S. = (B - x)(Cx - 2)

= BCx - 2B - Cx^2 + 2x

= -Cx^2 + (2 + BC)x - 2B

1) -2B = -3

B = -3 / -2

B = 1.5

2) -Cx^2 = 3x^2

-C = 3

C = -3

3) (2 + BC)x = Ax

[2 + 1.5(-3)] = A

2 - 4.5 = A

A = -2.5

希望以上解答可以幫到你~~~~~

2010-10-11 21:15:44 補充:
不要緊~

2a) 那條應該做對的~

2b) (x + y + z)^2

Let (y + z) be b.

(x + y + z)^2

= (x + b)^2

= x^2 + 2bx + b^2

= x^2 + 2x(y + z) + (y + z)^2

= x^2 + 2xy + 2xz + y^2 + 2yz + z^2

= x^2 + y^2 + z^2 + 2xy + 2xz + 2yz
參考: Yogi


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