平方差恆等式

2011-09-06 10:23 pm
利用平方差恆等式,分解為因式
1.98y^2 - 50x^2
2.18mp^2 - 2mq^2
3.16x^4 - 18
4.100 - (3x-7)^2
5.3(a-b)^ - 27(a+b)^2
6.a^2 - b^2+bc-ac

回答 (1)

2011-09-07 12:24 am
✔ 最佳答案
你好,我是STY,我的解答如下:

1) 98y^2 - 50x^2

= 2(49y^2 - 25x^2)

= 2[(7y)^2 - (5x)^2]

= 2(7y - 5x)(7y + 5x)

2) 18mp^2 - 2mq^2

= 2m(9p^2 - q^2)

= 2m[(3p)^2 - q^2]

= 2m(3p - q)(3p + q)

3) 16x^4 - 18

= 2(8x^4 - 9)

請檢查題目是吾打錯。

4) 100 - (3x - 7)^2

= 10^2 - (3x - 7)^2

= (10 - 3x + 7)(10 + 3x - 7)

= (- 3x + 17)(3x + 3)

= -(3x - 17)(3x + 3)

= -3(3x - 17)(x + 1)

5) 3(a - b)^2 - 27(a + b)^2

= 3[(a - b)^2 - 9(a + b)^2]

= 3{(a - b)^2 - [3(a + b)]^2}

= 3[(a - b)^2 - (3a + 3b)^2]

= 3(a - b - 3a - 3b)(a - b + 3a + 3b)

= 3(- 2a - 4b)(4a + 2b)

= -6(a + 2b)(4a + 2b)

= -12(a + 2b)(2a + b)

6) a^2 - b^2 + bc - ac

= (a - b)(a + b) + c(b - a)

= (a - b)(a + b) - c(a - b)

= (a - b)(a + b - c)

註:x^2 - y^2 = (x - y)(x + y)


BY STY~~~````

2011-09-06 17:45:39 補充:
3) 16x^4 - 81

= (4x^2)^2 - 9^2

= (4x^2 - 9)(4x^2 + 9)

= (2x - 3)(2x + 3)(4x^2 + 9)
參考: , STY!!!~```


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