2011 Λύσεις Σχ. β&I

4 i ªâùùúôóù ùô úôóôìûù ûâ

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4. i) ªÂÙ·ÙÚ¤ÔÓÙ·˜ ÙÔ˘˜ ·ÚÔÓÔÌ·ÛÙ¤˜ Û ÚËÙÔ‡˜ ¤¯Ô˘Ì ii) ∂›Ó·È ñ ñ Œ¯Ô˘Ì 5. i) ∞fi ÙÔ ˘ı·ÁfiÚÂÈÔ ıÂÒÚËÌ· ¤¯Ô˘ÌÂ μ° 2 = ∞μ 2 + ∞° 2 = · + ‚, ÔfiÙ ii) ™‡Ìʈӷ Ì ÙËÓ ÙÚÈÁˆÓÈ΋ ·ÓÈÛfiÙËÙ· ÈÛ¯‡ÂÈ μ° < ∞μ + ∞° Ô˘ ÛËÌ·›ÓÂÈ fiÙÈ iii) À„ÒÓÔ˘Ì ÛÙÔ ÙÂÙÚ¿ÁˆÓÔ Î·È ¤¯Ô˘Ì Ô˘ ÈÛ¯‡ÂÈ. ΔÔ “=” ÈÛ¯‡ÂÈ ·Ó Î·È ÌfiÓÔ ·Ó · = 0 ‹ ‚ = 0. · + ‚ ≤ · + ‚ + 2 · 0 ≤ 2 ·‚ , · + ‚ 2 · + 2 · + ‚ · + · + ‚ < · + ‚ . μ° = · + ‚. = 7 + 4 3 49 – 48 7 – 4 3 49 – 48 = 7 + 4 3 – 7 + 4 3 = 8 3. 1 2 – 3 2 1 2 + 3 2 = 1 7 – 4 3 1 7 + 4 3 2 + 3 2 = 4 + 4 3 + 3 = 7 + 4 3 ÔfiÙ 2 – 3 2 = 4 – 4 3 + 3 = 7 – 4 3 Î·È = 3 5 + 3 + 5 – 5 3 2 = 8 2 = 4. 3 5 3 + 5 5 + 3 = 3 5 + 3 5 – 3 + 5 5 3 5 – 3 = · + 1 · + 2 = · 2 + 1 + 2· · = · + 1 2 · · + 1 · 2 = · 2 + 1 · 2 + 2 · 1 · ∫∂º∞§∞π√ 2: √𠶃∞°ª∞Δπ∫√π ∞ƒπ£ª√π 26
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KEº∞§∞π√ 3 ∂•π™ø™∂π™ ¨ 3.1. ∂ÍÈÛÒÛÂȘ 1Ô˘ ‚·ıÌÔ‡ ∞ã √ª∞¢∞™ 1. i) 4x – 3(2x – 1) = 7x – 42 4x – 6x + 3 = 7x – 42 4x – 6x – 7x = –42 – 3 –9x = –45 x = 5. ÕÚ·, Ë Â͛ۈÛË ¤¯ÂÈ ÌÔÓ·‰È΋ χÛË, ÙËÓ x = 5. ii) 4(1 – 4x) – 5(x + 1) = x – 4 + 25 4 – 16x – 5x – 5 = x + 21 –21x – x = 21 + 1 –22x = 22 x= –1. ÕÚ·, Ë Â͛ۈÛË ¤¯ÂÈ ÌÔÓ·‰È΋ χÛË, ÙËÓ x = –1. iii) 30x – 20x = 15x – 12x – 49 30x – 20x – 15x + 12x = – 49 7x = –49 x = –7. ÕÚ·, Ë Â͛ۈÛË ¤¯ÂÈ ÌÔÓ·‰È΋ χÛË, ÙËÓ x = –7. iv) 1,2(x + 1) – 2,5 + 1,5x = 8,6 12(x + 1) – 25 + 15x = 86 12x + 12 – 25 + 15x = 86 27x = 99 x = ÕÚ·, Ë Â͛ۈÛË ¤¯ÂÈ ÌÔÓ·‰È΋ χÛË, ÙËÓ x = 11 3 . 2. i) 2(3x – 1) – 3(2x – 1) = 4 6x – 2 – 6x + 3 = 4 0x = 3. ÕÚ·, Ë Â͛ۈÛË Â›Ó·È ·‰‡Ó·ÙË. ii) ÕÚ·, Ë Â͛ۈÛË Â›Ó·È Ù·˘ÙfiÙËÙ·. 6x – 5 + x = –5 + 7x 0x = 0. 2x – 5 – x 3 = – 5 3 + 7x 3 3 2x – 3 5 – x 3 = 3 –5 3 + 3 7x 3 99 27 = 11 3 . x 2 x 3 = x 4 x 5 49 60 60 x 2 – 60 x 3 = 60 x 4 – 60 x 5 – 60 49 60 20 1 – 4x 5 – 20 x + 1 4 = 20 x – 4 20 + 20 5 4 1 – 4x 5 x + 1 4 = x – 4 20 + 5 4
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3. i) ñ ∞Ó Ï – 1 ≠ 0 Ï ≠ 1, ÙfiÙÂ Ë Â͛ۈÛË ¤¯ÂÈ ÌÔÓ·‰È΋ χÛË ÙËÓ ñ ∞Ó Ï = 1, ÙfiÙÂ Ë Â͛ۈÛË Á›ÓÂÙ·È 0x = 0 Î·È Â›Ó·È Ù·˘ÙfiÙËÙ·. ii) ñ ∞Ó Ï – 2 ≠ 0 Ï ≠ 2, ÙfiÙÂ Ë Â͛ۈÛË ¤¯ÂÈ ÌÔÓ·‰È΋ χÛË ÙËÓ ñ ∞Ó Ï = 2, ÙfiÙÂ Ë Â͛ۈÛË Á›ÓÂÙ·È 0x = 2 Î·È Â›Ó·È ·‰‡Ó·ÙË.
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