2011 Λύσεις Σχ. β&I

ªâ ùô ùô âúèôúèûìô ôìâ 6x 2 6x 1 2

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i) ∏ Â͛ۈÛË ÔÚ›˙ÂÙ·È ÁÈ· x ≠ –1 Î·È x ≠ 0. ªÂ ·˘ÙÔ‡˜ ÙÔ˘˜ ÂÚÈÔÚÈÛÌÔ‡˜ ¤¯Ô˘ÌÂ: 6x 2 + 6(x + 1) 2 = 13x(x + 1) 6x 2 + 6x 2 + 12x + 6 = 13x 2 + 13x x 2 + x – 6 = 0 Ë ÔÔ›· ¤¯ÂÈ Ú›˙˜ ÙÔ˘˜ ·ÚÈıÌÔ‡˜ 2 Î·È –3. ii) ∏ Â͛ۈÛË ÔÚ›˙ÂÙ·È ÁÈ· x ≠ 0 Î·È x ≠ 2. ªÂ ·˘ÙÔ‡˜ ÙÔ˘˜ ÂÚÈÔÚÈÛÌÔ‡˜ ¤¯Ô˘ÌÂ: 2x – 4 + 2x 2 – 3x + 2 – x 2 = 0 x 2 – x – 2 = 0. ∏ ÙÂÏÂ˘Ù·›· Â͛ۈÛË ¤¯ÂÈ Ú›˙˜ ÙÔ˘˜ ·ÚÈıÌÔ‡˜ 2 Î·È –1, ÔfiÙ ÏfiÁˆ ÙˆÓ ÂÚÈÔÚÈÛÌÒÓ ‰ÂÎÙ‹ Â›Ó·È ÌfiÓÔ Ë x = –1. 15. i) ∞Ó ı¤ÛÔ˘Ì x 2 = y Ë Â͛ۈÛË Á›ÓÂÙ·È y 2 + 6y – 40 = 0. ∞˘Ù‹ ¤¯ÂÈ Ú›˙˜ ÙȘ y 1 = 4 Î·È y 2 = –10. ∂Âȉ‹ y = x 2 ≥ 0, ‰ÂÎÙ‹ Â›Ó·È ÌfiÓÔ Ë y 1 = 4, ÔfiÙ ¤¯Ô˘Ì x 2 = 4 x = 2 ‹ x = –2. ∂Ô̤ӈ˜ ÔÈ Ú›˙˜ Ù˘ ·Ú¯È΋˜ Â͛ۈÛ˘ Â›Ó·È ÔÈ ·ÚÈıÌÔ› –2 Î·È 2. x(x – 2) 2 x + x(x – 2) 2x – 3 x – 2 + x(x – 2) 2 – x 2 x(x – 2) = 0 2 x + 2x – 3 x – 2 + 2 – x 2 x(x – 2) = 0 6x(x + 1) x x + 1 + 6x(x + 1) x + 1 x = 6x(x + 1) 13 6 x x + 1 + x + 1 x = 13 6 1, 3 – 5 2 Î·È 3 + 5 2 . 3 – 5 2 Î·È 3 + 5 2 . x + 1 x = 3 x 2 + 1 = 3x x 2 – 3x + 1 = 0 3.3. ∂ÍÈÛÒÛÂȘ 2Ô˘ ‚·ıÌÔ‡ 41
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ii) ∞Ó ı¤ÛÔ˘Ì x 2 = y Ë Â͛ۈÛË Á›ÓÂÙ·È 4y 2 + 11y – 3 = 0. ∞˘Ù‹ ¤¯ÂÈ Ú›- ˙˜ ÙȘ y 1 = –3 Î·È y 2 = 1 4 . ∂Âȉ‹ y = x 2 ≥ 0 ‰ÂÎÙ‹ Â›Ó·È ÌfiÓÔ Ë y 2 = 1 4 , ÔfiÙ ¤¯Ô˘Ì x 2 = 1 4 x = 1 2 ‹ x = – 1 2 . ∂Ô̤ӈ˜ ÔÈ Ú›˙˜ Ù˘ ·Ú¯È΋˜ Â͛ۈÛ˘ Â›Ó·È ÔÈ ·ÚÈıÌÔ› – 1 2 Î·È 1 2 . iii) ∞Ó ı¤ÛÔ˘Ì x 2 = y Ë Â͛ۈÛË Á›ÓÂÙ·È 2y 2 + 7y + 3 = 0. ∞˘Ù‹ ¤¯ÂÈ Ú›- ˙˜ ÙȘ y 1 = –3 Î·È y 2 = – 1 2 . ∂Âȉ‹ y = x 2 ≥ 0 η̛· ·fi ·˘Ù¤˜ ‰ÂÓ Â›- Ó·È ‰ÂÎÙ‹. ∂Ô̤ӈ˜ Ë ·Ú¯È΋ Â͛ۈÛË Â›Ó·È ·‰‡Ó·ÙË. ™¯fiÏÈÔ: ∂›Ó·È ÚÔÊ·Ó¤˜ fiÙÈ Ë Â͛ۈÛË Â›Ó·È ·‰‡Ó·ÙË, ·ÊÔ‡ 2x 4 + 7x 2 + 3 > 0 ÁÈ· οı x . μã √ª∞¢∞™ 1. i) ¢ = (–2· 3 ) 2 – 4· 2 4 – 1) = 4· 6 – 4· 6 + 4· 2 = 4· 2 . ii) √È Ú›˙˜ Ù˘ Â͛ۈÛ˘ Â›Ó·È 2. i) ∂›Ó·È ii) √È Ú›˙˜ Ù˘ Â͛ۈÛ˘ Â›Ó·È 3. i) ∏ Â͛ۈÛË ¤¯ÂÈ ‰ÈÏ‹ Ú›˙· ·Ó Î·È ÌfiÓÔ ·Ó ¢ = 0. ∂›Ó·È ¢ = (· – 9) 2 – 4 Ø 2(· 2 + 3· + 4) = · 2 – 18· + 81 – 8· 2 – 24· – 32 = –7· 2 – 42· + 49, ÔfiÙ x 2 = 5 – 2 2 – 1 2 = 4 – 2 2 2 = 2 – 2 . x 1 = 5 – 2 + 2 + 1 2 2 = 5 – 2 + 2 + 1 2 = 3 Î·È = 2 2 + 2 2 + 1 = 2 + 1 2 .
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