2011 Λύσεις Σχ. β&I

Iii ïï 1x ï 1 ñ ó ïï 1 0 ï 0 îè ï 1

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iii) Ï(Ï – 1)x = Ï –1 ñ ∞Ó Ï(Ï – 1) ≠ 0 Ï ≠ 0 Î·È Ï ≠ 1, ÙfiÙÂ Ë Â͛ۈÛË ¤¯ÂÈ ÌÔÓ·‰È΋ χ- ÛË ÙËÓ ñ ∞Ó Ï = 0 Ë Â͛ۈÛË Á›ÓÂÙ·È 0x = –1 Î·È Â›Ó·È ·‰‡Ó·ÙË. ñ ∞Ó Ï = 1 Ë Â͛ۈÛË Á›ÓÂÙ·È 0x = 0 Î·È Â›Ó·È Ù·˘ÙfiÙËÙ·. iv) Ï(Ï – 1)x = Ï 2 + Ï Ï(Ï – 1)x = Ï(Ï + 1). ñ ∞Ó Ï(Ï – 1) ≠ 0 Ï ≠ 0 Î·È Ï ≠ 1, ÙfiÙÂ Ë Â͛ۈÛË ¤¯ÂÈ ÌÔÓ·‰È΋ χ- ÛË ÙËÓ ñ ∞Ó Ï = 0, ÙfiÙÂ Ë Â͛ۈÛË Á›ÓÂÙ·È 0x = 0 Î·È Â›Ó·È Ù·˘ÙfiÙËÙ·. ñ ∞Ó Ï = 1, ÙfiÙÂ Ë Â͛ۈÛË Á›ÓÂÙ·È 0x = 2 Î·È Â›Ó·È ·‰‡Ó·ÙË. 4. ŒÛÙˆ ∞ª = x, ÙfiÙ ¢ª = 5 – x, ÔfiÙ i) ∏ ÈÛfiÙËÙ· ∂ 1 + ∂ 2 = ∂ 3 Â›Ó·È ÈÛÔ‰‡Ó·ÌË Ì ÙËÓ ÈÛfiÙËÙ· ·fi ÙËÓ ÔÔ›· ÚÔ·ÙÂÈ Ë Â͛ۈÛË ∂Ô̤ӈ˜ Ë ı¤ÛË ÙÔ˘ ª ÚÔÛ‰ÈÔÚ›˙ÂÙ·È ·fi ÙÔ Ì‹ÎÔ˜ ∞ª = 2,5, Â›Ó·È ‰ËÏ·‰‹ ÙÔ Ì¤ÛÔ ÙÔ˘ ∞¢. 30 – 6x + 10x = 40 4x = 10 x = 5 2 = 2,5. 3(5 – x) 2 + 5x 2 = (5 + 3)5 4 4 15 – 3x 2 + 4 5x 2 = 4 40 4 1 + ∂ 2 = (∞μ°¢) 2 1 = 3(5 – x) 2 Î·È 2 = x 5 2 . x = Ï(Ï + 1) Ï(Ï – 1) = Ï + 1 Ï – 1 . x = Ï – 1 Ï(Ï – 1) = 1 Ï . x = Ï Ï – 2 . x = Ï – 1 Ï – 1 = 1. ∫∂º∞§∞π√ 3: ∂•π™ø™∂π™ 28
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ii) ∏ ÈÛfiÙËÙ· ∂ 1 = ∂ 2 Â›Ó·È ÈÛÔ‰‡Ì·ÌË Ì ÙËÓ Â͛ۈÛË ∂Ô̤ӈ˜ Ë ı¤ÛË ÙÔ˘ ª ÚÔÛ‰ÈÔÚ›˙ÂÙ·È ·fi ÙÔ Ì‹ÎÔ˜ 5. ∞Ó ÙÔ ÔÛfi ÙˆÓ x ¢ÚÒ Î·Ù·Ù¤ıËΠÚÔ˜ 5%, ÙfiÙ ÙÔ ˘fiÏÔÈÔ ÔÛfi ÙˆÓ (4000 – x) ¢ÚÒ Î·Ù·Ù¤ıËΠÚÔ˜ 3%. – ΔÔ ÔÛfi ÙˆÓ x ¢ÚÒ ¤‰ˆÛ ÂÙ‹ÛÈÔ ÙfiÎÔ Â˘ÚÒ – ΔÔ ÔÛfi ÙˆÓ (4000 – x) ¢ÚÒ ¤‰ˆÛ ÂÙ‹ÛÈÔ ÙfiÎÔ Â˘ÚÒ. ∏ Â͛ۈÛË Ô˘ ·ÓÙÈÛÙÔȯ› ÛÙÔ Úfi‚ÏËÌ· Â›Ó·È 5x + 12.000 – 3x = 17.500 2x = 17.500 – 12.000 2x = 5.500 x = 2.750 ¢ÚÒ. ∂Ô̤ӈ˜ Ù· 2.750 ¢ÚÒ ÙÔΛÛÙËÎ·Ó ÚÔ˜ 5% Î·È Ù· ˘fiÏÔÈ· 1.250 ¢- ÚÒ ÙÔΛÛÙËÎ·Ó ÚÔ˜ 3%. 6. i) v = v 0 + ·t ·t = v – v 0 , ·ÊÔ‡ · ≠ 0. ii) ∞fi ÙËÓ ÙÂÏÂ˘Ù·›· ÈÛfiÙËÙ· ÚÔ·ÙÂÈ fiÙÈ R 2 – R ≠ 0, ·ÊÔ‡ ÙÔ ∂Ô̤ӈ˜ ¤¯Ô˘Ì 7. i) x 2 (x – 4) + 2x(x – 4) + (x – 4) = 0 (x – 4) (x 2 + 2x + 1) = 0 (x – 4) (x + 1) 2 = 0 x – 4 = 0 x + 1 = 0 x = 4 x = –1. ∂Ô̤ӈ˜ ÔÈ Ï‡ÛÂȘ Ù˘ Â͛ۈÛ˘ Â›Ó·È ÔÈ ·ÚÈıÌÔ› 4 Î·È –1. ii) (x – 2) 2 – (2 – x) (4 + x) = 0 (x – 2) 2 + (x – 2) (x + 4) = 0 (x – 2) [(x – 2) + (x + 4)] = 0 (x – 2) (2x + 2) = 0 x – 2 = 0 2x + 2 = 0 x = 2 x = –1.
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