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f.3 maths problems
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1. 16(p + q)2 - 25q2 = (4p + 4q)2 - (5q)2 = [(4p + 4q) + 5q] [(4p + 4q) - 5q] = (4p + 9q)(4p - q) 2. 9a3b - 12a2b2+ 4ab3 = ab(9a2 - 12ab + 4b2) = ab[(3a)2 - 2(3a)(2b) + (2b)2] = ab(3a - 2b)2 3. (a + b)3 + (a - b)3 = [(a + b) + (a - b)] [(a + b)2 - (a + b)(a - b) + (a - b)2] = (a + b + a - b)(a2 + 2ab + b2 - a2 + b2 + a2 - 2ab + b2) = 2a(a2 + 3b2) 2011-09-18 16:51:58 補充： Identities used : (u + v)2 = u2 + 2uv + v2 (u - v)2 = u2 - 2uv + v2 u2 - v2 = (u + v)(u - v) u3 + v3 = (u + v)(u2 - uv + v2) 2011-09-18 16:56:31 補充： Alternative methods: 1. 16(p + q)2 - 25q2 = 16p2 + 32pq + 16q2 - 25q2 = 16p2 + 32pq - 9q2 = (4p + 9q)(4p - q) 3. (a + b)3 + (a - b)3 = a3 + 3a2b + 3ab2 + b3 + a3 - 3a2b + 3ab2 - b3 = 2a3 + 6ab2 = 2a(a2 + 3b2)
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f.3 maths problems
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About Factorization 1.16(p+q)^2-25q^2 2,9a^3b-12a^2b^2+4^3 3.(a+b)^3+(a-b)^3 please help me:)thx alot!!最佳解答:
1. 16(p + q)2 - 25q2 = (4p + 4q)2 - (5q)2 = [(4p + 4q) + 5q] [(4p + 4q) - 5q] = (4p + 9q)(4p - q) 2. 9a3b - 12a2b2+ 4ab3 = ab(9a2 - 12ab + 4b2) = ab[(3a)2 - 2(3a)(2b) + (2b)2] = ab(3a - 2b)2 3. (a + b)3 + (a - b)3 = [(a + b) + (a - b)] [(a + b)2 - (a + b)(a - b) + (a - b)2] = (a + b + a - b)(a2 + 2ab + b2 - a2 + b2 + a2 - 2ab + b2) = 2a(a2 + 3b2) 2011-09-18 16:51:58 補充： Identities used : (u + v)2 = u2 + 2uv + v2 (u - v)2 = u2 - 2uv + v2 u2 - v2 = (u + v)(u - v) u3 + v3 = (u + v)(u2 - uv + v2) 2011-09-18 16:56:31 補充： Alternative methods: 1. 16(p + q)2 - 25q2 = 16p2 + 32pq + 16q2 - 25q2 = 16p2 + 32pq - 9q2 = (4p + 9q)(4p - q) 3. (a + b)3 + (a - b)3 = a3 + 3a2b + 3ab2 + b3 + a3 - 3a2b + 3ab2 - b3 = 2a3 + 6ab2 = 2a(a2 + 3b2)
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