F. 2 Maths -Identities (最佳解答取40分)

2006-12-16 4:18 am

1. (a)Factorize a^4-2a^2b^2+b^4
(b)If a/b+b/a=2, find the value of a^4-2a^2b^2+b^4.

2. It is given that y^2+1/y^2=7 and y=0, find the values of the following expressions.
(a)y+1/y
(b)(y-1/y)^2
(c)(y^2-1/y^2)^2

Remarks:

1) Please show the working steps clearly. (2 points)

2) Explain/state the working steps in details. (2 points)

3) Make sure your answer is correct. (Total points of each question =4 points)

-->5 X 4 X 2=40 points

Thanks a lot!

回答 (1)

用戶50266

2006-12-16 5:09 am

✔ 最佳答案

1(a) a^4-2a^2b^2+b^4
=(a^2-b^2)^2

(b) a/b+b/a=2
(a/b+b/a)^2 =4
(a^2/b^2)+(b^2/a^2)+2(a/b)(b/a) = 4
(a^2/b^2)+(b^2/a^2) = 2
(a^4+b^4)/(a^2b^2)=2
a^4+b^4=2a^2b^2
Therefore, a^4+b^4-2a^2b^2 = 0

2 Given that y^2+1/y^2=7

2(a) (y+1/y)^2 =y^2+2(y)(1/y)+(1/y^2 )
(y+1/y)^2 = 7+2
(y+1/y)^2 =9
y+1/y = 3 or -3 (rejected)
Therefore, y+1/y =3

(b) (y-1/y)^2
= (y+1/y)^2-4y(1/y)
= (3)^2 -4
= 9-4
=5

(c) Because (y-1/y)^2=5
y^2-2(y)(1/y)+(1/y^2) = 25
y^2-2+1/y^2 = 25
y^2+1/y^2 = 27
Also,
(y^2-1/y^2)^2 = (y^2+1/y^2)^2-4(y^2)(1/y^2)
= (27)^2-4
= 729-4
=725

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