我想問一條因式分解既問題,我要有解釋(中文)

此3題都可應用a^2+2ab+b^2=(a+b)^2或是a^2-2ab+b^2=(a-b)^2
1.x^2-8x+16
=(x-4)^2<----此是應用a^2-2ab+b^2=(a-b)^2,其中16是(4)^2=b^2,x^2=a^2
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2.x^2-4x+4
=(x-2)^2<----此是應用a^2-2ab+b^2=(a-b)^2,其中4是(2)^2=b^2,x^2=a^2
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3.9c^2-12c+4
=(3c-2)^2<----此是應用a^2-2ab+b^2=(a-b)^2,其中4是(2)^2=b^2,(3c)^2=a^2
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如有錯誤請指點

2009-10-17 20:42:36 補充：
上面d亂碼係

2009-10-17 20:44:30 補充：
←

恆等式(a-b)^2=a^2-2ab+b^2
1. x^2-8x+16
=x^2-2(x)(4)+4^2
=(x-4)^2
2.x^2-4x+4
=x^2-2(2)(x)+2^2
=(x-2)^2

3.9c^2-12c+4
=(3x)^2-2(3x)(2)+2^2
=(3x-2)^2

1. x^2 - 8x + 16
= x^2 - 4x - 4x + 16
= x(x - 4) - 4(x - 4)
= (x - 4)^2
2. x^2 - 4x + 4
= x^2 - 2x - 2x + 4
= x(x - 2) - 2(x - 2)
= (x - 2)^2
3. 9c^2 - 12c + 4
= 9c^2 - 6c - 6c + 4
= 3c(3c - 2) - 2(3c - 2)
= (3c - 2)^2