2011 Λύσεις Σχ. β&I

Óè 2 2 2 4 2 2 4 4 2 2 2 4 2 2 4 4 2 2 2 2 2 2 ñ

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∂›Ó·È ¢ = 2 + ‚ 2 ) 2 – 4· 2 2 = · 4 + ‚ 4 + 2· 2 2 – 4· 2 2 = · 4 + ‚ 4 – 2· 2 2 = (· 2 – ‚ 2 ) 2 ñ ∞Ó · ≠ ±‚ Ë Â͛ۈÛË ¤¯ÂÈ ‰‡Ô Ú›˙˜ ÙȘ ñ ∞Ó · = ‚ ‹ · = –‚ ÙfiÙÂ Ë Â͛ۈÛË ¤¯ÂÈ ‰ÈÏ‹ Ú›˙·, ÙËÓ 6. i) Œ¯Ô˘Ì ¢ = 4Ï 2 – 4 (–8) = 4Ï 2 + 32 > 0 ÁÈ· οıÂ Ï . ∞˘Ùfi ÛËÌ·›ÓÂÈ fiÙÈ Ë ÂÍ›- ÛˆÛË ¤¯ÂÈ Ú›˙˜ Ú·ÁÌ·ÙÈΤ˜ ÁÈ· οıÂ Ï . x = · 2 + ‚ 2 2‚ = 2 2‚ = · 2 = ± 2 = ‚. x 2 = · 2 + ‚ 2 – · 2 + ‚ 2 2‚ = 2‚ 2 2‚ = ‚. x 1 = · 2 + ‚ 2 + · 2 – ‚ 2 2‚ = 2 2‚ = · 2 Î·È x · + · x = · + · ·‚x x · + ·‚x · x = ·‚x · + ·‚x · x · + · x = · + · ·‚x x · + ·‚x · x = ·‚x · + ·‚x · x = ‚ x = · 2 . x = ‚ · ‚x = 1 · x = ‚ ‚x = · 2 ∫∂º∞§∞π√ 3: ∂•π™ø™∂π™ 44
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ii) ŒÛÙˆ x 1 , x 2 ÔÈ Ú›˙˜ Ù˘ Â͛ۈÛ˘ Ì x 2 = x 1 2 . ∞fi ÙÔ˘˜ Ù‡Ô˘˜ Vieta ¤¯Ô˘Ì ñ x 1 + x 2 = –2Ï x 1 + x 1 2 = –2Ï Î·È ñ x 1 Ø x 2 = –8 x 1 3 = –8 x 1 = –2, ÔfiÙ x 2 = (–2) 2 = 4. ΔfiÙ ¤¯Ô˘Ì –2 + 4 = –2Ï 2Ï = –2 Ï = –1. 7. ŒÛÙˆ x – 1, x, x + 1 ÙÚÂȘ ‰È·‰Ô¯ÈÎÔ› ·Î¤Ú·ÈÔÈ. √È ·ÚÈıÌÔ› ·˘ÙÔ› ·ÔÙÂ- ÏÔ‡Ó Ï¢ڤ˜ ÔÚıÔÁˆÓ›Ô˘ ÙÚÈÁÒÓÔ˘ ·Ó Î·È ÌfiÓÔ ·Ó ÈÛ¯‡ÂÈ (x + 1) 2 = x 2 + (x – 1) 2 x 2 + 2x + 1 = x 2 + x 2 – 2x + 1 x 2 – 4x = 0 x(x – 4) = 0 x = 4, ·ÊÔ‡ x ≠ 0 ˆ˜ ÏÂ˘Ú¿ ÙÚÈÁÒÓÔ˘. ∏ χÛË x = 4 Ù˘ Â͛ۈÛ˘ Â›Ó·È ÌÔÓ·‰È΋. ∂Ô̤ӈ˜ ˘¿Ú¯ÂÈ Ì›· ÌfiÓÔ ÙÚÈ¿‰· ‰È·‰Ô¯ÈÎÒÓ ·ÎÂÚ·›ˆÓ Ô˘ Â›Ó·È Ì‹ÎË Ï¢ÚÒÓ ÔÚıÔÁˆÓ›Ô˘ ÙÚÈÁÒ- ÓÔ˘. √È ·Î¤Ú·ÈÔÈ ·˘ÙÔ› Â›Ó·È ÔÈ 3, 4 Î·È 5. 8. ΔÔ ÂÌ‚·‰fiÓ ∂ 1 ÙÔ˘ ÛÙ·˘ÚÔ‡ ÚÔ·ÙÂÈ ·fi ÙÔ ¿ıÚÔÈÛÌ· ÙˆÓ ÂÌ‚·‰ÒÓ ÙˆÓ ‰‡Ô Ï¢ÎÒÓ ÏˆÚ›‰ˆÓ Ù˘ ÛËÌ·›·˜ ·fi ÙÔ ÔÔ›Ô fï˜ Ú¤ÂÈ Ó· ·Ê·ÈÚ¤ÛÔ˘Ì ÙÔ ÂÌ‚·‰fiÓ ÙÔ˘ ÎÔÈÓÔ‡ ÙÂÙÚ·ÁÒÓÔ˘ (√ªπ∑) ÏÂ˘Ú¿˜ d. ∂›Ó·È ‰ËÏ·‰‹ 1 = 3 Ø d + 4 Ø d – d 2 = 7d – d 2 ŒÛÙˆ ∂ 2 ÙÔ ÂÌ‚·‰fiÓ ÙÔ˘ ˘fiÏÔÈÔ˘ ̤- ÚÔ˘˜ Ù˘ ÛËÌ·›·˜. £· ÈÛ¯‡ÂÈ ∂ 1 = ∂ 2 ·Ó Î·È ÌfiÓÔ ·Ó ÙÔ ∂ 1 Â›Ó·È ›ÛÔ Ì ÙÔ ÌÈÛfi ÙÔ˘ ÂÌ‚·‰Ô‡ ÔÏfiÎÏËÚ˘ Ù˘ ÛËÌ·›- ·˜. ∂Ô̤ӈ˜ ¤¯Ô˘Ì 1 = ∂ 2 7d – d 2 = 3 Ø 4 2 d 2 – 7d + 6 = 0 d = 1 d = 6.
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