標題:

平方差恆等式

發問:

免費註冊體驗

 

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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 更新: 抱歉,第三題應該是 3.16x^4 - 81

最佳解答:

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)

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