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平方差恆等式
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你好,我是STY,我的解答如下: 1) 98y2 - 50x2 = 2(49y2 - 25x2) = 2[(7y)2 - (5x)2] = 2(7y - 5x)(7y + 5x) 2) 18mp2 - 2mq2 = 2m(9p2 - q2) = 2m[(3p)2 - q2] = 2m(3p - q)(3p + q) 3) 16x4 - 18 = 2(8x4 - 9) 請檢查題目是吾打錯。 4) 100 - (3x - 7)2 = 102 - (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) a2 - b2 + bc - ac = (a - b)(a + b) + c(b - a) = (a - b)(a + b) - c(a - b) = (a - b)(a + b - c) 註:x2 - y2 = (x - y)(x + y) BY STY```` 2011-09-06 17:45:39 補充: 3) 16x4 - 81 = (4x2)2 - 92 = (4x2 - 9)(4x2 + 9) = (2x - 3)(2x + 3)(4x2 + 9)
其他解答:
平方差恆等式
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利用平方差恆等式,分解為因式 1.98y2 - 50x2 2.18mp2 - 2mq2 3.16x4 - 18 4.100 - (3x-7)2 5.3(a-b) - 27(a+b)2 6.a2 - b2+bc-ac 更新: 抱歉,第三題應該是 3.16x4 - 81最佳解答:
你好,我是STY,我的解答如下: 1) 98y2 - 50x2 = 2(49y2 - 25x2) = 2[(7y)2 - (5x)2] = 2(7y - 5x)(7y + 5x) 2) 18mp2 - 2mq2 = 2m(9p2 - q2) = 2m[(3p)2 - q2] = 2m(3p - q)(3p + q) 3) 16x4 - 18 = 2(8x4 - 9) 請檢查題目是吾打錯。 4) 100 - (3x - 7)2 = 102 - (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) a2 - b2 + bc - ac = (a - b)(a + b) + c(b - a) = (a - b)(a + b) - c(a - b) = (a - b)(a + b - c) 註:x2 - y2 = (x - y)(x + y) BY STY```` 2011-09-06 17:45:39 補充: 3) 16x4 - 81 = (4x2)2 - 92 = (4x2 - 9)(4x2 + 9) = (2x - 3)(2x + 3)(4x2 + 9)
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