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# Ôìóˆ ë âíûˆûë âè ïûë ìfióô ó 0 0

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∂ÔÌ¤Óˆ˜ Ë ÂÍ›ÛˆÛË ¤¯ÂÈ Ï‡ÛË ÌfiÓÔ ·Ó · ≠ 0, ‚ ≠ 0 Î·È · ≠ ‚. x = ·‚ ‚ – · . x · x = 1 ‚x – ·x ·‚ = 1 (‚ – ·)x = ·‚. x = (· – ‚)(· + ‚) · – ‚ = · + ‚. x – · = x – ‚ · x = (· + ‚) 2 2(· + ‚) = · + ‚ 2 . ∫∂º∞§∞π√ 3: ∂•π™ø™∂π™ 34

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3. i) ™Ù· 200 ml ‰È¿Ï˘Ì· ÂÚÈ¤¯ÔÓÙ·È 30 ml Î·ı·Úfi ÔÈÓfiÓÂ˘Ì·. ∞Ó ÚÔ- Ûı¤ÛÔ˘ÌÂ x ml Î·ı·Úfi ÔÈÓfiÓÂ˘Ì· ÙfiÙÂ ÙÔ ‰È¿Ï˘Ì· Ô˘ ı· ÚÔÎ‡„ÂÈ ı· Â›Ó·È (200 + x) ml Î·È ı· ÂÚÈ¤¯ÂÈ (30 + x) ml Î·ı·Úfi ÔÈÓfiÓÂ˘Ì· ÔfiÙÂ ÚÔÎ‡ÙÂÈ Ë ÂÍ›ÛˆÛË ∂ÔÌ¤Óˆ˜ Ô Ê·ÚÌ·ÎÔÔÈfi˜ Ú¤ÂÈ Ó· ÚÔÛı¤ÛÂÈ 50 ml Î·ı·Úfi ÔÈÓfi- ÓÂ˘Ì·. 4. ŒÛÙˆ fiÙÈ x ÒÚÂ˜ ÌÂÙ¿ ÙËÓ ÚÔÛ¤Ú·ÛË Ù· ‰‡Ô ·˘ÙÔÎ›ÓËÙ· ı· ·¤¯Ô˘Ó ÌÂ- Ù·Í‡ ÙÔ˘˜ 1 km. ΔÔ ‰È¿ÛÙËÌ· Ô˘ ‰È·Ó‡ÂÈ ÙÔ ∞ ÛÙÈ˜ x ÒÚÂ˜ Â›Ó·È 100x ÂÓÒ ÙÔ ·ÓÙ›ÛÙÔÈ¯Ô ‰È¿ÛÙËÌ· ÁÈ· ÙÔ μ Â›Ó·È 120x. ŒÙÛÈ ¤¯Ô˘ÌÂ ÙËÓ ÂÍ›ÛˆÛË 120x – 100x = 1 20x = 1 x = 1 20 ÒÚÂ˜, ÔfiÙÂ x = 1 20 Ø 60 = 3 ÏÂÙ¿. √fiÙÂ Ù· ·˘ÙÔÎ›ÓËÙ· ı· ·¤¯Ô˘Ó 1km ÙÚ›· ÏÂÙ¿ ÌÂÙ¿ ÙËÓ ÚÔÛ¤Ú·ÛË. 5. ∏ ÂÍ›ÛˆÛË ·˘Ù‹ Â›Ó·È ÔÚÈÛÌ¤ÓË ÁÈ· x ≠ · Î·È x ≠ –·. ªÂ ·˘ÙÔ‡˜ ÙÔ˘˜ Â- ÚÈÔÚÈÛÌÔ‡˜ ¤¯Ô˘ÌÂ: (x + ·) 2 = x 2 x + · = x x + · = –x 0x = · 2x = –·. ñ ∞Ó · = 0, ÙfiÙÂ Ë ÂÍ›ÛˆÛË ¤¯ÂÈ ˆ˜ Ï‡ÛË Î¿ıÂ ·ÚÈıÌfi x ≠ 0. ñ ∞Ó · ≠ 0, ÙfiÙÂ Ë ÂÍ›ÛˆÛË ¤¯ÂÈ ÌÔÓ·‰ÈÎ‹ Ï‡ÛË ÙÔÓ ·ÚÈıÌfi . 6. ∏ ÂÍ›ÛˆÛË ·˘Ù‹ Â›Ó·È ÔÚÈÛÌ¤ÓË ÁÈ· x ≠ 2. ªÂ ·˘Ùfi ÙÔÓ ÂÚÈÔÚÈÛÌfi ¤¯Ô˘ÌÂ: x 3 – 8 = x 3 – 2x 2 + 4x – 8 2x 2 – 4x = 0 2x(x – 2) = 0 x = 0 x = 2. ∞fi ÙÈ˜ ÙÈÌ¤˜ ·˘Ù¤˜ ‰ÂÎÙ‹ Â›Ó·È ÌfiÓÔ Ë x = 0 ∂ÔÌ¤Óˆ˜ Ë ÂÍ›ÛˆÛË ¤¯ÂÈ ÌÔÓ·‰ÈÎ‹ Ï‡ÛË, ÙÔÓ ·ÚÈıÌfi x = 0. x 3 – 8 x – 2 = x 2 + 4 x = –· 2 x + · x – · = x 2 x 2 – · 2 x + · x – · = x 2 (x + ·)(x – ·) x = 3400 68 x = 50. 3000 + 100x = 6400 + 32x 68x = 3400 30 + x 200 + x = 32 100 100(30 + x) = 32(200 + x) 3.1. ∂ÍÈÛÒÛÂÈ˜ 1Ô˘ ‚·ıÌÔ‡ 35
7. |2|x| – 1| = 3 2|x| – 1 = 3 2|x| – 1 = –3 2|x| = 4 2|x| = –2. ∏ ‰Â‡ÙÂÚË Â›Ó·È ·‰‡Ó·ÙË ÔfiÙÂ ¤¯Ô˘ÌÂ 2|x| = 4 |x| = 2 x = –2 x = 2. ∂ÔÌ¤Óˆ˜ ÔÈ Ï‡ÛÂÈ˜ ÙË˜ ÂÍ›ÛˆÛË˜ Â›Ó·È ÔÈ ·ÚÈıÌÔ› –2 Î·È 2.

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