{"id":1325,"date":"2025-12-22T14:43:28","date_gmt":"2025-12-22T13:43:28","guid":{"rendered":"http:\/\/mymatma.pl\/?p=1325"},"modified":"2025-12-22T14:43:32","modified_gmt":"2025-12-22T13:43:32","slug":"zadanie-132","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=1325","title":{"rendered":"Zadanie 132"},"content":{"rendered":"\n<p><strong>zadanie 15 \u2013 maj 2025 (zadanie otwarte) (3pkt)<\/strong><\/p>\n\n\n\n<p>Wyznacz warto\u015b\u0107 m , dla kt\u00f3rej trzywyrazowy ci\u0105g (2m+11,m<sup>2<\/sup>+3,5\u2212m) jest arytmetyczny i malej\u0105cy.<\/p>\n\n\n\n<div class=\"wp-block-columns is-layout-flex wp-container-core-columns-is-layout-28f84493 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n    <div class=\"premium-wrapper-odp\" id=\"odp_69e22ba36aa05\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22ba36aa05\"\n                style=\"margin-top:10px; padding:12px 20px; background:#28a745; color:#fff; border-radius:25px; border:none; cursor:pointer; font-weight:bold;\">\n            Poka\u017c odpowied\u017a\n        <\/button>\n\n        <div class=\"premium-content-odp\" style=\"display:none; padding:15px; border:1px solid #28a745; margin-top:10px; background:#e6ffe6; border-radius:25px;\">\n            \nPrawid\u0142owa odpowied\u017a to m = 1.\n        <\/div>\n    <\/div>\n    \n<\/div>\n\n\n\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n    <div class=\"premium-wrapper-wyjasnienie\" id=\"wyjasnienie_69e22ba36aa0f\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22ba36aa0f\"\n                style=\"margin-top:10px; padding:12px 20px; background:#28a745; color:#fff; border-radius:25px; border:none; cursor:pointer; font-weight:bold;\">\n            Poka\u017c wyja\u015bnienie\n        <\/button>\n\n        <div class=\"premium-content-wyjasnienie\" style=\"display:none; padding:15px; border:1px solid #28a745; margin-top:10px; background:#e6ffe6; border-radius:25px;\">\n            \n<h3>Wyja\u015bnienie<\/h3>\n\n<p>W zadaniu dany jest trzywyrazowy ci\u0105g:<\/p>\n\n<p style=\"text-align:center;\">\n(2m + 11), &nbsp; (m\u00b2 + 3), &nbsp; (5 \u2212 m)\n<\/p>\n\n<p>Aby ci\u0105g by\u0142 arytmetyczny, r\u00f3\u017cnica mi\u0119dzy kolejnymi wyrazami musi by\u0107 taka sama. Zapisujemy warunek:<\/p>\n\n<p style=\"text-align:center;\">\n(m\u00b2 + 3) \u2212 (2m + 11) = (5 \u2212 m) \u2212 (m\u00b2 + 3)\n<\/p>\n\n<h4>1. Upraszczamy lew\u0105 stron\u0119<\/h4>\n\n<p style=\"text-align:center;\">\nm\u00b2 + 3 \u2212 2m \u2212 11 = m\u00b2 \u2212 2m \u2212 8\n<\/p>\n\n<h4>2. Upraszczamy praw\u0105 stron\u0119<\/h4>\n\n<p style=\"text-align:center;\">\n5 \u2212 m \u2212 m\u00b2 \u2212 3 = \u2212m\u00b2 \u2212 m + 2\n<\/p>\n\n<h4>3. Otrzymujemy r\u00f3wnanie<\/h4>\n\n<p style=\"text-align:center;\">\nm\u00b2 \u2212 2m \u2212 8 = \u2212m\u00b2 \u2212 m + 2\n<\/p>\n\n<p>Przenosimy wszystko na jedn\u0105 stron\u0119:<\/p>\n\n<p style=\"text-align:center;\">\n2m\u00b2 \u2212 m \u2212 10 = 0\n<\/p>\n\n<h4>4. Rozwi\u0105zujemy r\u00f3wnanie kwadratowe<\/h4>\n\n<p>\u0394 = (\u22121)\u00b2 \u2212 4\u00b72\u00b7(\u221210) = 1 + 80 = 81<\/p>\n\n<p style=\"text-align:center;\">\nm = (1 \u00b1 9) \/ 4\n<\/p>\n\n<p>Otrzymujemy dwa rozwi\u0105zania:<\/p>\n\n<p style=\"text-align:center;\">\nm\u2081 = 10\/4 = 2.5  \n<br>\nm\u2082 = \u22128\/4 = \u22122\n<\/p>\n\n<hr>\n\n<h4>5. Sprawdzamy, dla kt\u00f3rego m ci\u0105g jest malej\u0105cy<\/h4>\n\n<p>R\u00f3\u017cnica ci\u0105gu to:<\/p>\n\n<p style=\"text-align:center;\">\nr = (m\u00b2 + 3) \u2212 (2m + 11) = m\u00b2 \u2212 2m \u2212 8\n<\/p>\n\n<p>Aby ci\u0105g by\u0142 malej\u0105cy, musi by\u0107:<\/p>\n\n<p style=\"text-align:center;\">\nr &lt; 0\n<\/p>\n\n<h4>Sprawdzamy oba rozwi\u0105zania:<\/h4>\n\n<ul>\n  <li>Dla m = 2.5: r = 2.5\u00b2 \u2212 2\u00b72.5 \u2212 8 = 6.25 \u2212 5 \u2212 8 = \u22126.75 &lt; 0 \u2714\ufe0f<\/li>\n  <li>Dla m = \u22122: r = (\u22122)\u00b2 \u2212 2\u00b7(\u22122) \u2212 8 = 4 + 4 \u2212 8 = 0 \u274c (ci\u0105g nie jest malej\u0105cy)<\/li>\n<\/ul>\n\n<h3>Ostatecznie:<\/h3>\n\n<p style=\"font-size:1.2em;\">\n<strong>m = 2.5<\/strong>\n<\/p>\n\n<p>To jedyna warto\u015b\u0107, dla kt\u00f3rej ci\u0105g jest jednocze\u015bnie arytmetyczny i malej\u0105cy.<\/p>\n\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 15 \u2013 maj 2025 (zadanie otwarte) (3pkt) Wyznacz warto\u015b\u0107 m , dla kt\u00f3rej trzywyrazowy ci\u0105g (2m+11,m2+3,5\u2212m) jest arytmetyczny i malej\u0105cy.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"saved_in_kubio":false,"_uag_custom_page_level_css":"","_themeisle_gutenberg_block_has_review":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[25,152,204,81],"tags":[23,154,205],"class_list":["post-1325","post","type-post","status-publish","format-standard","hentry","category-ciag-arytmetyczny","category-maj-2025","category-rozne","category-zadania-maturalne","tag-ciag-arytmetyczny","tag-maj-2025","tag-rozne"],"jetpack_featured_media_url":"","uagb_featured_image_src":{"full":false,"thumbnail":false,"medium":false,"medium_large":false,"large":false,"1536x1536":false,"2048x2048":false,"kubio-fullhd":false,"woocommerce_thumbnail":false,"woocommerce_single":false,"woocommerce_gallery_thumbnail":false},"uagb_author_info":{"display_name":"adminmymatma","author_link":"https:\/\/mymatma.pl\/?author=1"},"uagb_comment_info":0,"uagb_excerpt":"zadanie 15 \u2013 maj 2025 (zadanie otwarte) (3pkt) Wyznacz warto\u015b\u0107 m , dla kt\u00f3rej trzywyrazowy ci\u0105g (2m+11,m2+3,5\u2212m) jest arytmetyczny i malej\u0105cy.","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1325","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1325"}],"version-history":[{"count":1,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1325\/revisions"}],"predecessor-version":[{"id":1326,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1325\/revisions\/1326"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1325"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1325"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}