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(1a) x2 - 8x = 0 (x2 - 8x + 16) - 16 = 0 (x - 4)2 - 16 = 0 (x - 4)2 = 16 x - 4 = 4 or -4 x = 8 or 0 (1b) x2 + 10x = -16 x2 + 10x + 25 = 9 (x + 5)2 = 9 x + 5 = 3 or -3 x = -2 or -8 (11a) Sub x = 2, we have y = 5, i.e. 5 = 4a +2b + 1 2a + b = 2 ... (1) Sub x = -1, we have y = 11, i.e. 11 = a - b + 1 a - b = 10 ... (2) Solving we have a = 4 and b = -6 (b) y = 4x2 - 6x + 1 = 4(x2 - 3x/2 + 9/16) + 1 - 9/4 = 4(x - 3/4)2 - 5/4 So when x = 3/4, y is minimum = -5/4 (c) Vertex: (3/4, -5/4) Axis of symmetry: x = 3/4 (12a) BC = x + 2, AB = 118 - x - (x + 2) = 116 - 2x (b) Area is give by: (x + 2)(116 - 2x) = 2(x + 2)(58 - x) = 2(-x2 + 56x + 116) = 2(-x2 + 56x - 784) + 1800 = -2(x - 28)2 + 1800 So when x = 28, the area is maximum = 1800 (c) AB = 60 m and BC = 20 m (2) Let the numbers be x and x + 2, then x(x + 2) = 575 x2 + 2x - 575 = 0 (x - 23)(x + 25) = 0 x = 23 or -25 (rejected) So the numbers are 23 and 25. (9a) ∠ECD = ∠EAB and ∠EDC = ∠EBA (ext. ∠s of cyclic. quad.) So △CED~△AEB (AAA) (9b) Since △CED~△AEB, CE/AE = ED/EB 5/(4 + AD) = 4/12 AD = 11 (10a) △HAD~△HCB (10b) Since △HAD~△HCB, HC/HA = HB/HD 6/12 = 7/(6 + DC) DC = 8
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好唔抵-_- 2008-11-25 22:14:07 補充： 唉，我唔答啦－?－ 都唔知要答邊題，下次打出黎嘛ＸＰ －－－－－－－－－ １ａ ｘ２－８ｘ＝０ （ｘ－４）２－４２＝０ （ｘ－４）＝正負４ ｘ＝８ｏｒｘ＝０ １ｂ ｘ２＋１０ｘ＝－１６ （ｘ＋５）２＝－１６＋（５）２ （ｘ＋５）＝正負３ ｘ＝－８ｏｒ－２
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[IMG]http://i293.photobucket.com/albums/mm67/zaza520/0001-59.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0002-60.jpg[/IMG][IMG]http://i293.photobucket.com/albums/mm67/zaza520/0003-45.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0005-44.jpg[/IMG]最佳解答:
(1a) x2 - 8x = 0 (x2 - 8x + 16) - 16 = 0 (x - 4)2 - 16 = 0 (x - 4)2 = 16 x - 4 = 4 or -4 x = 8 or 0 (1b) x2 + 10x = -16 x2 + 10x + 25 = 9 (x + 5)2 = 9 x + 5 = 3 or -3 x = -2 or -8 (11a) Sub x = 2, we have y = 5, i.e. 5 = 4a +2b + 1 2a + b = 2 ... (1) Sub x = -1, we have y = 11, i.e. 11 = a - b + 1 a - b = 10 ... (2) Solving we have a = 4 and b = -6 (b) y = 4x2 - 6x + 1 = 4(x2 - 3x/2 + 9/16) + 1 - 9/4 = 4(x - 3/4)2 - 5/4 So when x = 3/4, y is minimum = -5/4 (c) Vertex: (3/4, -5/4) Axis of symmetry: x = 3/4 (12a) BC = x + 2, AB = 118 - x - (x + 2) = 116 - 2x (b) Area is give by: (x + 2)(116 - 2x) = 2(x + 2)(58 - x) = 2(-x2 + 56x + 116) = 2(-x2 + 56x - 784) + 1800 = -2(x - 28)2 + 1800 So when x = 28, the area is maximum = 1800 (c) AB = 60 m and BC = 20 m (2) Let the numbers be x and x + 2, then x(x + 2) = 575 x2 + 2x - 575 = 0 (x - 23)(x + 25) = 0 x = 23 or -25 (rejected) So the numbers are 23 and 25. (9a) ∠ECD = ∠EAB and ∠EDC = ∠EBA (ext. ∠s of cyclic. quad.) So △CED~△AEB (AAA) (9b) Since △CED~△AEB, CE/AE = ED/EB 5/(4 + AD) = 4/12 AD = 11 (10a) △HAD~△HCB (10b) Since △HAD~△HCB, HC/HA = HB/HD 6/12 = 7/(6 + DC) DC = 8
其他解答:
好唔抵-_- 2008-11-25 22:14:07 補充： 唉，我唔答啦－?－ 都唔知要答邊題，下次打出黎嘛ＸＰ －－－－－－－－－ １ａ ｘ２－８ｘ＝０ （ｘ－４）２－４２＝０ （ｘ－４）＝正負４ ｘ＝８ｏｒｘ＝０ １ｂ ｘ２＋１０ｘ＝－１６ （ｘ＋５）２＝－１６＋（５）２ （ｘ＋５）＝正負３ ｘ＝－８ｏｒ－２
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