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Bu kavramlarla ilgili temel düzeyde bazı problemler sunulmuştur. Derslerin pdf dosyalarını indirmek için...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/2834505/90a2ba3518597a50658449375bf6f0c9/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/HwKDlwAAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"1","reqid":"1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL","summary":{"fullTextUrl":"/int_search_summary?data=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3D3BZXx3UHKYE","linkTemplate":"/video/preview/4681322367785623643?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Logaritma -1 Lise Matematik Dersleri","related_orig_text":"LG Math","related_porno":false,"related_less_3m_off":true,"client":"d2d","no_cnt":1,"related_src":"serp","related":"{\"porno\":false,\"vfp\":1,\"orig_text\":\"LG Math\",\"url\":\"http:\\/\\/www.youtube.com\\/watch?v=3BZXx3UHKYE\",\"src\":\"serp\",\"rvb\":\"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-AgD2-AYJDgb8AQEM7v_2_v0A9AYC_wcC_wAA6QgBAwAAAP8OAAn6AAAA_-71CP39AAANBAH-BAAAABL-APj_AAAAB_zw9wABAAD1_gMFAwAAABAO_f3_AAAACAP6BwAAAAAKAAABAAAAAATkAf8BAAAAIAAtuojVOzgTQAlITlACKoQCEAAa8AF_9DX-wvnD_73t9v_H_8sApSj2APw_yACpzgMB2zK2AavoDgDmSwj_LBDsAKk1IwEb8N7-8570ADTNEf8X6hgA1ggsARD78QJFMR8AC8flAedCE_3uAhAD_brXABNI9_waAfH4-AG7_xL_ygQt-jkBAev6_x_iCP_c1Bj86PQRCMkXyP35R-IE_-7m_s8FKgIX_hQJ-DQL-vfl0v0A9xP8JOUU8BIv6wVFKvsCCh3rB9fn6wbf9_H-9AUS-9nu5QHCwAX_4eL-A_bJEwsC9hry4yILA1sxEQL_3QUDAA8H-_L55v_g2f0B4vfrAtbAD_YgAC1Ne_E6OBNACUhhUAIqcxAAGmA59AA38AHk2A719sjzOTu7oOTKGcrW__bc_-4L7ekcD_UB_RMAKecP3aIAAAA_38gV8wDge6rlLC4CERqPnwgR13_6HTbcJA79t-wd9hgI3f3vDy4A69exH1IVnC5LEzYgAC32thM7OBNACUhvUAIqrwYQDBqgBgAAIMEAADDCAAAAQgAAIMEAANjBAAA0QgAAfEIAAChCAAC4wQAAsEEAACBBAABQwQAAgL8AAAAAAAAwQgAAQEIAAGDBAABQwQAAgEIAAEBAAADQwQAAgEAAAPjBAADgQQAA4MEAAKDBAADgQAAAMEEAALBBAADIQQAAWMIAABhCAABkwgAAgD8AAL7CAAAgQQAAcEIAAIhCAAC4wQAAQMAAANjBAABAwQAAbEIAAMjBAACowQAAgMEAAKBAAACAwAAAwEAAAAhCAACgwAAAsMEAAIhBAABgwgAANEIAACDBAADgwQAA-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_AACgwAAAiEEAAKDAAABgQgAAssIAANDBAAAcwgAAKEIAABzCAACYQQAAoMAAAKjCAACwwQAAoEAAAEDAAABAQgAAoEIAACDBAABIwgAAQEEAAFBBAAB4wgAAHMIAAJzCAADAwAAAqMEAAKBBAAAgQQAAaMIAAJjBAAAEwgAAEMEAAKBBAAAQQgAAmMIAAATCAAAcQgAAsMEAAIDBAABcQgAA0EEAAMBBAABgwgAANEIAAFhCAAD4wQAAbEIAAFjCIAA4E0AJSHVQASqPAhAAGoACAAAQvQAAhr4AAJg9AABkvgAAEL0AAHw-AADgPAAAX78AAKa-AAB8PgAABL4AAJq-AAC4PQAAQLwAAKi9AACSvgAAHD4AAMi9AACKPgAAMT8AAH8_AAAwvQAAJD4AAHS-AADiPgAA6L0AAJ4-AAD4PQAAsr4AABA9AABsPgAALL4AAGw-AACmPgAAwj4AAOi9AADIPQAA2L0AAIq-AAAEvgAATL4AACy-AACYPQAA2r4AACQ-AAC2vgAAoDwAAIC7AADOvgAAqr4AAFy-AAABvwAAgDsAANg9AACWvgAAUD0AAFM_AADqvgAANL4AAKY-AAAcPgAA0j4AACQ-AABAvCAAOBNACUh8UAEqjwIQARqAAgAAgj4AAOI-AAB0vgAAR78AANa-AACoPQAADz8AAKC8AACSPgAA6D0AAIY-AABsvgAABL4AAES-AAD4PQAAQLwAAES-AADePgAA0r4AAMY-AACgPAAA0r4AAOi9AAAkvgAA4LwAAFA9AABsvgAAPD4AAFy-AADIvQAAmD0AAEA8AAB8vgAABD4AAKg9AAD4vQAAcD0AAHA9AACGvgAA2L0AAI4-AADIPQAAQLwAAOg9AAA0vgAAUD0AAH-_AABkPgAAyD0AAHC9AAA8vgAAjj4AAFS-AAAsPgAA2D0AAGw-AAAMvgAAMD0AALg9AACgvAAAoDwAAEC8AABMPgAAFD4gADgTQAlIfFABMAk4AUoAYABoAA,,\"}","related_url":"http://www.youtube.com/watch?v=3BZXx3UHKYE","parent-reqid":"1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL","related_vfp":1,"relatedVideo":"yes"},"cwidth":560,"cheight":360,"cratio":1.55555,"dups":["4681322367785623643"],"episode":0,"season":0,"isEmbedOnly":false,"greenHost":"YouTube","hasTranslation":true,"contentTypeId":null,"censored":false,"videoContentId":"499361978"},"2495513503164239506":{"videoId":"2495513503164239506","docid":"34-8-10-Z0515A45C64434026","description":"Bu derste sinüs, kosinüs, tanjant, kotanjant, sekant, kosekant trigonometrik fonksiyonlarının birim çember yardımıyla tanımları verilmiştir. 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İlk bölümde onluk logaritma (taban 10) ve doğal logaritma (taban e) tanımları yapılmakta, e sayısının yaklaşık 2,71 değerine sahip olduğu açıklanmaktadır. İkinci bölümde ise logaritma özellikleri (logaritma a tabanında 1'in değerinin sıfır olduğu, e tabanında 1'in değerinin 0 olduğu, 10 tabanında 10'un değerinin 1 olduğu) ve taban değiştirme kuralı örneklerle anlatılmaktadır.","Videoda logaritma değerlerinin hesaplanması, logaritma ifadelerinin birbirine dönüştürülmesi ve logaritma özellikleri kullanılarak problemlerin çözülmesi gibi konular işlenmektedir. Taban değiştirme kuralının bir başka videoda detaylı olarak anlatılacağı bilgisi de paylaşılmaktadır."]},"endTime":772,"title":"Logaritma Dersi: Onluk ve Doğal Logaritma","beginTime":0}],"fullResult":[{"index":0,"title":"Onluk ve Doğal Logaritma","list":{"type":"unordered","items":["Logaritma derslerinin üçüncü videosunda onluk logaritma ve doğal logaritma olarak adlandırılan iki özel tabana sahip logaritma anlatılacak.","Onluk logaritma, tabanı on olan logaritmadır ve herhangi bir taban yazılmamışsa log x sembolünden on tabanındaki logaritmayı anlamalıyız.","lnx sembolü, e tabanındaki logaritmayı gösterir ve e sayısı yaklaşık olarak 2,71 değerine sahiptir."]},"beginTime":0,"endTime":49,"href":"/video/preview/12232487152877619659?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=0&ask_summarization=1"},{"index":1,"title":"E Sayısının Hesaplanması","list":{"type":"unordered","items":["E sayısı, limit n sonsuza giderken (1+1/n)^n ifadesinin yaklaşık 2,71 değerine yaklaştığı görülmüştür.","n=1 için 2, n=2 için 2,25, n=3 için 2,71'e giderek yaklaşıldığı gösterilmiştir.","E sayısının hesaplanması konunun dışında olduğu için detaylı olarak ele alınmamıştır."]},"beginTime":49,"endTime":148,"href":"/video/preview/12232487152877619659?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=49&ask_summarization=1"},{"index":2,"title":"Onluk Logaritma Örnekleri","list":{"type":"unordered","items":["Log 2 yaklaşık olarak 0,31 ve log 3 yaklaşık olarak 0,47 olarak verilmiştir.","18.000 logaritması hesaplanırken, 18.000 = 18 × 1000 şeklinde parçalanarak log 18 + log 1000 = 2 × log 3 + 3 = 0,94 + 0,31 + 3 = 4,25 olarak bulunmuştur.","Log 75 hesaplanırken, 75 = 25 × 3 şeklinde parçalanarak log 25 + log 3 = 2 × log 5 + log 3 = 2 × (0,69 + 0,47) şeklinde hesaplanmıştır."]},"beginTime":148,"endTime":434,"href":"/video/preview/12232487152877619659?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=148&ask_summarization=1"},{"index":3,"title":"Logaritma Özellikleri ve Örnekler","list":{"type":"unordered","items":["Log 10 = 2 + log 5 şeklinde parçalanabilir ve log 5 = x verildiğinde log 2 = 1 - 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Konuşmacı, 2017 Nisan ayında karşılaştığı bir sayılar teorisi problemi ve kendi geliştirdiği çözümü anlatmaktadır.","Video, 383 üzeri 101 sayısının son üç basamağının nasıl hesaplanacağını Euler teoremi ve mod parçalama yöntemi kullanarak göstermektedir. Konuşmacı önce Euler teoreminin temel prensiplerini açıklar, ardından mod 1000'de 383 üzeri 101'in son üç basamağını bulmak için modu 125 ve 8'e ayırarak Euler teoremini uygular. Bu yöntem sayesinde daha hızlı ve doğru bir sonuç elde edilir. Video, modun yüksek olduğu durumlarda Euler teoremini nasıl uygulayabileceğimizi gösteren bir öğretici içerik sunmaktadır."]},"endTime":559,"title":"Sayılar Teorisinde Euler Teoremi ve Mod Parçalama Yöntemi","beginTime":0}],"fullResult":[{"index":0,"title":"Sayılar Teorisinden Bir Problem","list":{"type":"unordered","items":["Videoda sayılar teorisinden bir problem tanıtılacak ve 2017 Nisan ayında yabancı bir site ile karşılaşılmış.","Problemin binom açılımından faydalanarak hesaplamaları uğraştırıcı bir çözümü vardı, ancak konuşmacı Çin Kalan Teoremi içeren daha keskin ve hızlı bir çözüm tasarlamış.","Problem: 383 üzeri 101 sayısının son üç basamağı kaçtır?"]},"beginTime":0,"endTime":56,"href":"/video/preview/14418189321355735451?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=0&ask_summarization=1"},{"index":1,"title":"Euler Teoremi ve Kullanımı","list":{"type":"unordered","items":["Euler teoremi: n bir doğal sayı, a bir tam sayı ve n ile a aralarında asal ise, a üzeri (n-1) değeri mod n'de 1'e eşittir.","Problemin çözümünde mod 1000'de işlem yapmak gerekiyor ve Euler fonksiyonu (fi) hesaplanması gerekiyor.","fi(1000) değeri 400 olarak hesaplanıyor çünkü 1000'in asal çarpanları 2³×5³ şeklinde ve fi(1000) = (2³-2²)×(5³-5²) = 4×100 = 400 olarak bulunuyor."]},"beginTime":56,"endTime":198,"href":"/video/preview/14418189321355735451?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=56&ask_summarization=1"},{"index":2,"title":"Modu Parçalama Yöntemi","list":{"type":"unordered","items":["Euler teoremi doğrudan uygulanamadığı için modu parçalama yöntemi tercih ediliyor.","383 sayısının hangi kuvveti mod 125'te 1'e eşit olduğunu ve mod 8'de 1'e eşit olduğunu bulmak gerekiyor.","mod 125 için fi(125) = 5³-5² = 100 olarak hesaplanıyor ve 383 üzeri 100 değeri mod 125'te 1'e 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olabilir."]},"beginTime":224,"endTime":413,"href":"/video/preview/2798305862790384497?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=224&ask_summarization=1"},{"index":4,"title":"Fonksiyonların Kontrolü","list":{"type":"unordered","items":["f(x) = x fonksiyonu verilen denklemi sağlar.","f(x) = 1/x fonksiyonu da verilen denklemi sağlar.","Verilen fonksiyonel denklemi sağlayan iki fonksiyon bulunmuştur."]},"beginTime":413,"endTime":612,"href":"/video/preview/2798305862790384497?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=413&ask_summarization=1"},{"index":5,"title":"Farklı Fonksiyonların İncelenmesi","list":{"type":"unordered","items":["f(x) fonksiyonunun sadece x veya 1/x değerlerini alması gerektiği belirtilmiştir.","f(a) = a ve f(b) = 1/b şeklinde farklı fonksiyonlar da incelenebilir.","Bu durumda x=a, w=b ve y, z = √ab değerleri denklemde yerine konularak yeni bir 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derste belirsiz integral ve diferansiyel kavramlarının tanımları verilip temel düzeyde bazı problemler sunulmuştur. Ders dokümanlarının bağlantısı https://drive.google.com/open?id=1HO9...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/1780987/e2fadd3e5f376dcd1bfbfc308da644b9/564x318_1"},"target":"_self","position":"17","reqid":"1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=videoid:5210915938642105570","teaser":[{"list":{"type":"unordered","items":["Bu video, bir matematik dersi formatında integral konusuna giriş yapan bir eğitim içeriğidir. Eğitmen, belirsiz integralin temel kavramlarını ve uygulamalarını anlatmaktadır.","Video, belirsiz integralin tanımıyla başlayıp, integralin sembolleri (integral sembolü, diferansiyel dx, integrad) ve belirsiz integral sabiti (C) hakkında bilgiler vermektedir. Eğitmen, temel türev alma kurallarını kullanarak belirsiz integral alma yöntemlerini örneklerle açıklamakta ve diferansiyel kavramını tanıtmaktadır. Video, integral alma kurallarını kullanmadan soruları çözmeyi gösteren örneklerle devam eder ve bir sonraki derste temel integral teoremleri ve formüllerin anlatılacağı bilgisiyle sonlanır."]},"endTime":610,"title":"İntegral Konusu Eğitim Videosu","beginTime":0}],"fullResult":[{"index":0,"title":"İntegralin Tanımı","list":{"type":"unordered","items":["İntegral konusuna giriş yapılıyor ve ilk olarak belirsiz integral tanıtılacak.","Belirsiz integral, türevi verilen f fonksiyonunun bulunması işlemidir ve küçük f fonksiyonunun integrali f(x) + C sembolü ile gösterilir.","C sayısına belirsiz integral sabiti denir."]},"beginTime":0,"endTime":32,"href":"/video/preview/5210915938642105570?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=0&ask_summarization=1"},{"index":1,"title":"İntegral Örnekleri","list":{"type":"unordered","items":["Türevi 2x olan f fonksiyonu için, x², x² + 2, x² - 5 gibi farklı fonksiyonlar olabilir çünkü sabitlerin türevi sıfırdır.","İntegral sembolü, dx (x'e göre diferansiyel), f(x) (integrale alınacak fonksiyon) ve C (belirsiz integral sabiti) olarak tanımlanır.","İntegral alma işlemi, türevi verilen fonksiyonu bulmak için yapılır."]},"beginTime":32,"endTime":223,"href":"/video/preview/5210915938642105570?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=32&ask_summarization=1"},{"index":2,"title":"Diferansiyel Kavramı","list":{"type":"unordered","items":["Diferansiyel, fonksiyonun türevinin yanına dx çarpanı getirilerek elde edilir.","dy/dx = f(x) ifadesi, f(x) dx şeklinde yazılabilir ve bu ifadeye f(x) diferansiyeli denir.","Diferansiyel alma işlemi, fonksiyonun türevini alıp yanına dx çarpanı eklemektir."]},"beginTime":223,"endTime":322,"href":"/video/preview/5210915938642105570?parent-reqid=1769621945730998-3078963379391317601-balancer-l7leveler-kubr-yp-vla-130-BAL&text=LG+Math&t=223&ask_summarization=1"},{"index":3,"title":"İntegral ve 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Eğitmen, trigonometri denklemlerinin çözüm yöntemlerini adım adım anlatmaktadır.","Video, trigonometri denklemlerinin çözüm yöntemlerini göstermektedir. İlk olarak sinüs ve kosinüs fonksiyonlarının iki kare farkı kullanılarak çarpanlara ayrılması, ardından tanjant fonksiyonunun kullanımı anlatılmaktadır. Daha sonra sinüs ve kosinüs denklemlerinin çözüm kümesinin eleman sayısı ve köklerinin toplamı gibi sorular çözülmektedir. Her soru için farklı çözüm yöntemleri gösterilmekte ve yarım açı formülleri, iki kat açı formülleri gibi trigonometrik formüller kullanılmaktadır."]},"endTime":911,"title":"Trigonometri Denklemleri Çözüm Dersi","beginTime":0}],"fullResult":[{"index":0,"title":"Trigonometrik Denklemlerin Çözümü","list":{"type":"unordered","items":["Sinüs üzeri dört x eksi kosinüs üzeri dört x ifadesi iki kare farkı olarak çarpanlara ayrılarak sinüs kare x artı üç çarpı kosinüs kare x şeklinde yazılabilir.","Bu ifade pozitif bir değerdir ve bir'den büyük veya bir'e eşit olabilir, 90 derecede bir olur.","Denklemin çözümü için sinüs kare x eksi üç, kosinüs kare x eksi bir şeklinde iki kare farkı uygulanır ve sinüs x eşittir kök üç kosinüs x veya sinüs x eşittir eksi kök üç kosinüs x 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