Form 2 Identities questions

2008-01-08 2:45 am
(Answer the following questions by using the identities of the different of two squares and the perfect square.)

1. (a) Expand (1+1/x)(1-1/x)

(b) Hence, show that 12/11 x 10/11 = (112 – 1)/112

2. (a) Expand (x+b)2

(b) Using the result of (a), expand (x+y+z)2

回答 (2)

2008-01-08 3:05 am
✔ 最佳答案
1. (a) Expand (1+1/x)(1-1/x)
(1+1/x)(1-1/x)
Let a=1/x

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

12/11 x 10/11
= (1 + 1/11) (1 - 1/11)
Let x = 11;
[(x^2) - 1] / (x^2)
=[11^2 - 1)] / (11^2)

2. (a) Expand (x+b)2

(x+b)^2
= (x+b)(x+b)
= x^2 + 2xb + b^2

(b) Using the result of (a), expand (x+y+z)2

Let b=y+z

(x+y+z)^2
= (x + b) ^2
= x^2 + 2xb + 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 + 2(xy + xz + yz)
2008-01-08 3:06 am
1a(1+1/x)(1-1/x)
=1^2-1/x^2 <-----------(a+b)(a-b)=a^2-b^2
=1-1/x^2

1b(1+1/11)(1-1/11)
=1-1/121
=(121-1)/121
=(11^2-1)11^2


2a(x+b)^2 <-----------(a+b)^2=a^2+2ab+b^2
=x^2+2bz+b^2

2b(x+y+z)^2
=x^2+2(y+Z)x+(y+z)^2
=x^2+2xy+2xz+y^2+2yz+z^2
=x^2+y^2+z^2+2xy+2xz+2yz


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