代數補項及拆項法問題

用戶2053

2009-10-18 11:02 pm

✔ 最佳答案

1. x^2 - 4xy + 4y^2- 6x +12y + 8

= (x^2 - 4xy + 4y^2) - (6x - 12y) + 8

= (x - 2y)^2 - 6(x - 2y) + 8

= (x - 2y)^2 + (- 2 + - 4)(x - 2y) + (-2)(-4)

=(x - 2y - 2)(x - 2y - 4)

2. （x +y ）（x +y +2xy）+（xy + 1）（xy - 1）

= (x + y)^2 + 2xy(x + y) + (xy + 1)(xy - 1)

= (x + y)^2 + [(xy + 1) + (xy - 1)] (x + y) + (xy + 1)(xy - 1)

= (x + y + xy + 1)(x + y + xy - 1)

3. （x^2 + 3x - 1）（x^2 + 3x - 2）- 6

設 (x^2 + 3x - 1) = y :

原式 = y(y - 1) - 6

= y^2 - y - 6

= y^2 + (2 + -3)y + (2)(- 3)

= (y + 2)(y - 3)

原式 = (x^2 + 3x - 1 + 2)(x^2 + 3x - 1 - 3)

=(x^2 + 3x + 1)(x^2 + 3x - 4)

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[匿名]

2009-10-29 5:38 am

1. x^2 - 4xy + 4y^2- 6x +12y + 8

= (x^2 - 4xy + 4y^2) - (6x - 12y) + 8

= (x - 2y)^2 - 6(x - 2y) + 8

= (x - 2y)^2 + (- 2 + - 4)(x - 2y) + (-2)(-4)

=(x - 2y - 2)(x - 2y - 4)

2. （x +y ）（x +y +2xy）+（xy + 1）（xy - 1）

= (x + y)^2 + 2xy(x + y) + (xy + 1)(xy - 1)

= (x + y)^2 + [(xy + 1) + (xy - 1)] (x + y) + (xy + 1)(xy - 1)

= (x + y + xy + 1)(x + y + xy - 1)

3. （x^2 + 3x - 1）（x^2 + 3x - 2）- 6

設 (x^2 + 3x - 1) = y :

原式 = y(y - 1) - 6

= y^2 - y - 6

= y^2 + (2 + -3)y + (2)(- 3)

= (y + 2)(y - 3)

原式 = (x^2 + 3x - 1 + 2)(x^2 + 3x - 1 - 3)

=(x^2 + 3x + 1)(x^2 + 3x - 4)

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