標題:
發問:
1)(a) Prove that (m^2-1)^2+(2m)^2=(m^2+1)^2 is an identity. (b)Using (a) , find positive integers a and b such that a^2+b^2=65^2
最佳解答:
1) (a) Prove that (m2-1)2 + (2m)2 = (m2+1)2 is an identity. LHS = (m2-1)2 + (2m)2 = m? - 2m2 + 1 + 4m2 = m? + 2m2 + 1 = (m2+1)2 = RHS (b) Using (a), find positive integers a and b such that a2 + b2 = 652 Consider m2 + 1 = 65 m2 = 64 m = 8 or -8 (rejected as the question requires positive integer) m2 - 1 = 63 2m = 16 Therefore, we have 632 + 162 = 652. 2013-09-15 15:01:11 補充： hei, 你可以睇下, 這個是得出 Pythagorian triple的一個方法： http://en.wikipedia.org/wiki/Pythagorean_triple 以本題為例，即是可以有一直角三角形，邊長為16, 63, 65。
其他解答:
LHS=(m^2-1)^2+(2m)^2 =m^4-m^2-m^2+1+4m^2 =m^4+2m^2+1 =(m^2+1)^2=RHS Let m^2+1=65 a=m^2-1 b=2m m=8 a=63 b=16|||||1(a) (m^2 - 1)^2 + (2m)^2 = m^4 - 2m^2 + 1 + 4m^2 = m^4 + 2m^2 + 1 =(m^2 + 1)^2 So, (m^2 - 1)^2 + (2m)^2 = (m^2 + 1)^2 is an identity (b) Sub. m = 8, a = m^2 - 1 = 63 and b = 2m = 16
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F2 Maths Hard Question發問:
1)(a) Prove that (m^2-1)^2+(2m)^2=(m^2+1)^2 is an identity. (b)Using (a) , find positive integers a and b such that a^2+b^2=65^2
最佳解答:
1) (a) Prove that (m2-1)2 + (2m)2 = (m2+1)2 is an identity. LHS = (m2-1)2 + (2m)2 = m? - 2m2 + 1 + 4m2 = m? + 2m2 + 1 = (m2+1)2 = RHS (b) Using (a), find positive integers a and b such that a2 + b2 = 652 Consider m2 + 1 = 65 m2 = 64 m = 8 or -8 (rejected as the question requires positive integer) m2 - 1 = 63 2m = 16 Therefore, we have 632 + 162 = 652. 2013-09-15 15:01:11 補充： hei, 你可以睇下, 這個是得出 Pythagorian triple的一個方法： http://en.wikipedia.org/wiki/Pythagorean_triple 以本題為例，即是可以有一直角三角形，邊長為16, 63, 65。
其他解答:
LHS=(m^2-1)^2+(2m)^2 =m^4-m^2-m^2+1+4m^2 =m^4+2m^2+1 =(m^2+1)^2=RHS Let m^2+1=65 a=m^2-1 b=2m m=8 a=63 b=16|||||1(a) (m^2 - 1)^2 + (2m)^2 = m^4 - 2m^2 + 1 + 4m^2 = m^4 + 2m^2 + 1 =(m^2 + 1)^2 So, (m^2 - 1)^2 + (2m)^2 = (m^2 + 1)^2 is an identity (b) Sub. m = 8, a = m^2 - 1 = 63 and b = 2m = 16
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