\u0130\u00e7inde bilinmeyen bulunan ve bilinmeyenin baz\u0131 de\u011ferleri i\u00e7in do\u011fru olan e\u015fitsizliklere denklem denir.
\n Denklemi sa\u011flayan bilinmeyenin de\u011ferine o denklemin k\u00f6k\u00fc ya da k\u00f6kleri denir. Denklemin k\u00f6k\u00fcn\u00fc veya k\u00f6klerini bulmak i\u00e7in yap\u0131lan i\u015fleme denklemi \u00e7\u00f6zme; k\u00f6k veya k\u00f6klerin olu\u015fturdu\u011fu k\u00fcmeye ise \u00e7\u00f6z\u00fcm k\u00fcmesi denir.
\n Denklem; i\u00e7indeki bilinmeyen say\u0131s\u0131 ve bilinmeyenin \u00fcss\u00fcne g\u00f6re adland\u0131r\u0131l\u0131r. <\/p>\n
O HALDE;
\n 5x \u2013 5 = 15, y + 2 = 6 a\u00e7\u0131k \u00f6nermeleri bir bilinmeyenli birinci dereceden bir denklemdir.
\n 2x + y = 9 a\u00e7\u0131k \u00f6nermesi iki bilinmeyenli birinci dereceden bir denklemdir.
\n x + y + z = 4 a\u00e7\u0131k \u00f6nermesi \u00fc\u00e7 bilinmeyenli birinci dereceden bir denklemdir.
\n x\u00b2 – 9 = 16 a\u00e7\u0131k \u00f6nermesi ikinci dereceden bir bilinmeyenli denklemdir.<\/p>\n
\u0130\u00e7inde bir tane bilinmeyeni bulunan ve \u00fcss\u00fc bir olan denklemlere birinci dereceden bir bilinmeyenli denklemler denir.<\/p>\n
Genel olarak; a,b,c \u0404 R ve a \u2260 0 olmak \u00fczere ax + b = c \u015feklinde g\u00f6sterilen denklemlere birinci dereceden bir bilinmeyenli denklem denir. <\/p>\n
DENKLEM \u00c7\u00d6Z\u00dcM\u00dcNDE B\u0130L\u0130NMES\u0130 GEREKEN \u00d6ZELL\u0130KLER<\/p>\n
1. Bir e\u015fitli\u011fin her iki yan\u0131na ayn\u0131 reel say\u0131
\neklenirse, e\u015fitlik bozulmaz. Bu \u00f6zeli\u011fe; e\u015fitli\u011fin toplama kural\u0131 denir.<\/p>\n
2. Bir e\u015fitli\u011fin her iki yan\u0131 da s\u0131f\u0131rdan farkl\u0131
\nayn\u0131 reel say\u0131yla \u00e7arp\u0131l\u0131rsa, e\u015fitlik bozulmaz. Bu \u00f6zeli\u011fe; e\u015fitli\u011fin \u00e7arpma kural\u0131 denir.<\/p>\n
3. Bir e\u015fitli\u011fin her iki yan\u0131 da s\u0131f\u0131rdan farkl\u0131
\nayn\u0131 reel say\u0131ya b\u00f6l\u00fcn\u00fcrse, e\u015fitlik bozulmaz. Bu \u00f6zeli\u011fe; e\u015fitli\u011fin b\u00f6lme kural\u0131 denir.<\/p>\n
4. Bir denklemde herhangi bir terimi e\u015fitli\u011fin
\nbir taraf\u0131ndan di\u011fer taraf\u0131na ge\u00e7irerek i\u015flem yapmak gerekiyorsa; ge\u00e7irilen terimin i\u015fareti de\u011fi\u015ftirilir.<\/p>\n
Pratik \u00c7\u00f6z\u00fcm
\n Bir denklemi pratik \u00e7\u00f6zmek i\u00e7in ;
\n Bilinmeyenler e\u015fitli\u011fin bir yan\u0131nda, bilinenler e\u015fitli\u011fin di\u011fer yan\u0131nda toplan\u0131r. E\u015fitli\u011fin bir yan\u0131ndan di\u011fer yan\u0131na ge\u00e7en terimin i\u015fareti de\u011fi\u015fir.
\n Her iki yanda toplama \u00e7\u0131karma i\u015flemleri yap\u0131l\u0131r ve her iki yan bilinmeyenin katsay\u0131s\u0131na b\u00f6l\u00fcnerek bilinmeyen yaln\u0131z b\u0131rak\u0131l\u0131r. Denklem \u00e7\u00f6z\u00fclm\u00fc\u015f olur.<\/p>\n
\u00d6RNEKLER<\/p>\n
1. x + 6 = 10 denkleminin \u00e7\u00f6z\u00fcm k\u00fcmesini
\nbulal\u0131m:<\/p>\n
\u00c7\u00f6z\u00fcm:
\nx + 6 = 10 denkleminde (+6) n\u0131n toplama
\ni\u015flemine g\u00f6re ters eleman\u0131 olan (-6), e\u015fitli\u011fin her iki yan\u0131na eklenirse e\u015fitlik bozulmaz.<\/p>\n
Buna g\u00f6re; x + 6 = 10
\n x + 6 + (-6) = 10 + (-6)
\n x + 0 = 4
\n x = 4 olur.
\n \u00c7 = {4} olur.<\/p>\n
Verilen bir denklemin \u00e7\u00f6z\u00fcm\u00fcn\u00fcn do\u011fru yap\u0131l\u0131p yap\u0131lmad\u0131\u011f\u0131n\u0131n ara\u015ft\u0131r\u0131lmas\u0131na, denklemin sa\u011flamas\u0131 denir.<\/p>\n
Bulunan k\u00f6k, denklemde yerine yaz\u0131larak denklemin sa\u011flamas\u0131 yap\u0131l\u0131r b\u00f6ylece bulunan k\u00f6k\u00fcn do\u011frulu\u011fu kontrol edilir.<\/p>\n
4 say\u0131s\u0131n\u0131n x + 6 = 10 denklemini sa\u011flay\u0131p sa\u011flamad\u0131\u011f\u0131n\u0131 kontrol edelim:<\/p>\n
x = 4 i\u00e7in x + 6 = 10
\n 4 + 6 =10
\n 10 = 10 oldu\u011fundan
\n \u00e7\u00f6z\u00fcm do\u011frudur.
\n x + 6 = 10
\n x = 10 \u2013 6
\n x = 4 ve \u00c7 = {4} t\u00fcr. <\/p>\n
Demek ki; her iki \u015fekilde yap\u0131lan \u00e7\u00f6z\u00fcm, ayn\u0131 eleman\u0131 veren \u00e7\u00f6z\u00fcm k\u00fcmesidir.<\/p>\n
2. Verilen denklem parantezli olursa; a\u015fa\u011f\u0131da yap\u0131ld\u0131\u011f\u0131 gibi, \u00f6nce da\u011f\u0131lma \u00f6zeli\u011fi uygulanarak parantezler kald\u0131r\u0131l\u0131r. Sonra da i\u00e7erisinde bilinmeyeni olan terimler e\u015fitli\u011fin bir taraf\u0131na, \u00f6teki terimler de di\u011fer taraf\u0131na ge\u00e7irilir. Gerekli i\u015flemler yap\u0131larak denklem \u00e7\u00f6z\u00fcl\u00fcr. <\/p>\n
2.(x + 3) + 7 = 25 \u2013 2.( x – 2 )
\n\u00d6nce, \u00e7arpma i\u015fleminin toplama ve \u00e7\u0131karma i\u015flemleri \u00fczerine da\u011f\u0131lma \u00f6zeliklerini uygulayal\u0131m<\/p>\n
\u00c7\u00f6z\u00fcm:<\/p>\n
2.(x + 3) + 7 = 25 \u2013 2.( x – 2 )
\n 2x + 6 + 7 = 25 \u2013 2x + 4
\n 2x + 13 = -2x + 29
\n 2x + 2x = 29 \u2013 13
\n 4x = 16
\n x = 16 : 4
\n x = 4 ve \u00c7 = { 4 } olur.<\/p>\n
3. Verilen denklem kesirli olursa, \u00e7\u00f6z\u00fcm\u00fc i\u00e7in \u00f6nce paydalar e\u015fitlenir. Denklem paydadan kurtar\u0131l\u0131r. Bunun i\u00e7in, e\u015fitli\u011fin iki yan\u0131n\u0131 ortak payda ile \u00e7arpmak gerekir. Sonra da \u00f6rnek \u00e7\u00f6z\u00fcmlerde belirtilen kurallara g\u00f6re denklem \u00e7\u00f6z\u00fcl\u00fcr.
\n 3.(x\u20132) _ 2\u2013x _ _ x _ 5 denkleminin \u00e7\u00f6z\u00fcm
\n 4 2 \u00af 5 2 k\u00fcmesini bulal\u0131m:<\/p>\n
\u00c7\u00f6z\u00fcm:
\nPaydalar\u0131 e\u015fitlersek:<\/p>\n
3.( x- 2) \u2013 2.( 2 \u2013 x ) \u2013 4x _ x – 10
\n 4 \u00af 4<\/p>\n
3x \u2013 6 \u2013 4 + 2x \u2013 4x =x \u2013 10
\n 3x + 2x \u2013 4x \u2013 x = -10 + 6 + 4
\n 5x – 5x = -10 + 10
\n 0.x = 0
\nBu e\u015fitlik b\u00fct\u00fcn reel say\u0131lar i\u00e7in ge\u00e7erli oldu\u011fundan verilen denklemin \u00e7\u00f6z\u00fcm k\u00fcmesi \u00c7=R d\u0131r.
\n4. 5 say\u0131s\u0131n\u0131n, 2x \u2013 6 = 3 denkleminin k\u00f6k\u00fc olup olmad\u0131\u011f\u0131n\u0131 ara\u015ft\u0131ral\u0131m:<\/p>\n
\u00c7\u00f6z\u00fcm:
\n x = 5 i\u00e7in 2x \u2013 6 = 3
\n 2 . 5 \u2013 6 = 3
\n 10 \u2013 6 = 3
\n 4 \u2260 3 olur<\/p>\n","protected":false},"excerpt":{"rendered":"
\u0130\u00e7inde bilinmeyen bulunan ve bilinmeyenin baz\u0131 de\u011ferleri i\u00e7in do\u011fru olan e\u015fitsizliklere denklem denir. Denklemi sa\u011flayan bilinmeyenin de\u011ferine o denklemin k\u00f6k\u00fc ya da k\u00f6kleri denir. Denklemin k\u00f6k\u00fcn\u00fc veya k\u00f6klerini bulmak i\u00e7in yap\u0131lan i\u015fleme denklemi \u00e7\u00f6zme; k\u00f6k veya k\u00f6klerin olu\u015fturdu\u011fu k\u00fcmeye ise \u00e7\u00f6z\u00fcm k\u00fcmesi denir. Denklem; i\u00e7indeki bilinmeyen say\u0131s\u0131 ve bilinmeyenin \u00fcss\u00fcne g\u00f6re adland\u0131r\u0131l\u0131r. O HALDE; 5x …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1407,1403],"tags":[2780,2781,2782],"class_list":["post-933","post","type-post","status-publish","format-standard","hentry","category-fen-ve-teknoloji-odevleri","category-odevler","tag-1-dereceden-1-bilinmeyenli-denklemler","tag-ortak-payda","tag-reel-sayilar"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/933","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/comments?post=933"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/933\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=933"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=933"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=933"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}