TANIM: : a bir reel ger\u00e7el say\u0131 ve n\u00ceZ+ olsun. a.a.a…a=an olacak \u015fekilde, n tane a\u2019n\u0131n \u00e7arp\u0131m\u0131 olan an e \u00fcsl\u00fc ifadeler denir.
\n \u00d6rnek\/ a) 3.3.3.3=34 b) c)
\n UYARI :8 a bir reel say\u0131 ve n\u00ceZ+ olmak \u00fczere a+a+a+…+a = n.a oldu\u011fu i\u00e7in an ile n.a ifadeleri birbirine kar\u0131\u015ft\u0131r\u0131lmamal\u0131d\u0131r. Yani an \u00b9 n.a d\u0131r. <\/p>\n
\u00d6rnek \/ 2+2+2+2+2 = 5.2 olup ayn\u0131 \u015fekilde 2.2.2.2.2 = 25 oldu\u011funa dikkat edilmelidir.<\/p>\n
Not : 1-) a\u00b90 olmak \u015fart\u0131yla a0 = 1 dir.
\n 2-) 00 = ifadesi tan\u0131ms\u0131zd\u0131r.
\n 3-) 1n = 1 dir (n\u00ceIR)<\/p>\n
\u00d6rnek\/ a) 80 =1 b) c) ( bu gibi \u00f6rneklerde parantez i\u00e7inin bilinmesi gerekir.) d) 115 =1 e) 1-15 = 1 f) <\/p>\n
—————\u00dcss\u00fcn \u00dcss\u00fc——————–
\n Tan\u0131m8 Bir \u00fcsl\u00fc ifadenin \u00fcss\u00fc \u00fcslerin \u00e7arp\u0131m\u0131na e\u015fittir. Kural <\/p>\n
\u00d6rnek\/ a) ( 52)3 = 52.3 =56 b) c) <\/p>\n
Not \/ 1- \u015feklindeki bir yaz\u0131l\u0131m ifadesi yanl\u0131\u015ft\u0131r. \u00c7\u00fcnk\u00fc n say\u0131s\u0131n\u0131n; m nin \u00fcss\u00fcm\u00fc yoksa am nin \u00fcss\u00fcm\u00fc oldu\u011fu belli de\u011fildir.
\n 2- dir. \u00dcslerin parantezlerle neyin \u00fcss\u00fc oldu\u011fu belirtilmelidir.
\n \u00d6rnek \/ oldu\u011funu g\u00f6sterin.<\/p>\n
a) = 32.3 =36 = 729
\n b) = 32.2.2 = 38 =6561<\/p>\n
Sonu\u00e7 : a ve b de\u011ferlerinden yukar\u0131da verilen e\u015fitsizli\u011fin dorulu\u011fu g\u00f6r\u00fclm\u00fc\u015ft\u00fcr.<\/p>\n
————————-Negatif \u00dcs Kavram\u0131—————–
\n Tan\u0131m 8 a bir reel say\u0131 olmak \u00fczere dir. Benzer \u015fekilde a\u00b90 ve b\u00b90 olmak \u00fczere
\n \u00d6rnek \/ 5-1 + 5-2 = ?=
\n \u00d6rnek \/ <\/p>\n
————————Bir Reel Say\u0131n\u0131n \u00dcss\u00fc——————-<\/p>\n
Tanm8 Pozitif say\u0131lar\u0131n b\u00fct\u00fcn kuvvetleri pozitiftir. Kural a > 0 Þ an > 0 d\u0131r.
\n \u00d6rnek \/ a) 42 = 16 > 0 b) 4-2 = c) 40 = 1 > 0
\n Tan\u0131m : 1- Negatif say\u0131lar\u0131n \u00c7ift Kuvvetleri Pozitiftir. Kural a < 0 ve n bir \u00e7ift say\u0131 ise an > 0<\/p>\n
Tan\u0131m : 2- Negatif say\u0131lar\u0131n Tek Kuvvetleri Negatiftir.Kural a < 0 ve n bir tek say\u0131 ise an < 0\n \u00d6rnek \/ 1- (-4)2 = 16 > 0
\n \u00d6rnek \/ 2- (-4)3 = -64 < 0\n\n Not 8 a > 0 ve n bir \u00e7ift say\u0131 ise (-a)n \u00b9 -an e\u015fitsizli\u011fi do\u011frudur.<\/p>\n
\u00d6rnek \/ 1- (-2)4 \u00b9 -24 \u00c7\u00fcnk\u00fc (-2)4 = (+16) ve \u201324 = -2.2.2.2= -16
\n \u00d6rnek \/ 2- (-5)3 + (-53) = (- 125) + (-125) = (-250)
\n \u00d6rnek \/ 3- (-5)4 + (-54) = (+625) + (-625) = 0
\n \u00d6rnek \/ 4- (-3)3 + (-52) + (-4)2 = (-27) + (-25) + (+16) = (-36)<\/p>\n
———————\u00dcsl\u00fc \u0130fadelerde D\u00f6rt \u0130\u015flem——————-<\/p>\n
1- Toplama ve \u00c7\u0131karma \u0130\u015flemi<\/p>\n
Tan\u0131m : \u00dcsl\u00fc ifadelerde toplama ve \u00e7\u0131karma i\u015fleminin yap\u0131labilmesi i\u00e7in benzer terimlerin \u00fcs ve tabanlar\u0131n\u0131n ayn\u0131 olmas\u0131 gerekir<\/p>\n
Kural :4 a.Xn b.Xn = (a b).Xn<\/p>\n
\u00d6rnek \/ 1- 5.103 + 2.103 = (5+2).103
\n \u00d6rnek \/ 1- 5.103 – 2.103 = (5-2).103<\/p>\n
Not8 m \u00b9 n ise am an i\u015flemi bu haliyle yap\u0131lamaz.
\n \u00d6rnek \/ 105 + 104 = i\u015fleminde 5 4 olup d\u00fczenleme yaparak i\u015flem tamamlan\u0131r.
\n 1.105 = 10.104
\n Burdan 10.104 + 1.104 = (10+1). 104
\n \u00d6rnek \/ 55 + 54 = 5.54 + 54 = (5+1). 54<\/p>\n
2- \u00c7arpma ve B\u00f6lme \u0130\u015flemi<\/p>\n
Tan\u0131m: Bir \u00fcsl\u00fc ifadede \u00c7arpma ve B\u00f6lme \u0130\u015fleminin yap\u0131labilmesi i\u00e7in benzer terimlerin tabanlar\u0131n\u0131n ay\u0131n\u0131 olmas\u0131 gerekir.<\/p>\n
Kural 8\/ 1- (a.Xm) .(b.Xn) = (a.b).Xm+n
\n Kural 8 2- (a.Xm) \u00b8 (b.Xn) = (a\u00b8b).Xm-n veya
\n \u00d6rnek \/ (2.52 ) . (3.54) = 2.3.52+4 =6.56
\n \u00d6rnek \/ (8.36) \u00b8 (4.32) =
\n \u00d6rnek \/
\n \u00d6rnek \/ 15a = 3a-2 oldu\u011funa g\u00f6re 5a n\u0131n de\u011ferini bulal\u0131m.
\n 15a = 3a-2 = (3.5)a = \u015feklinde yaz\u0131l\u0131rsa
\n 15a = 3a-2 = (3.5)a =
\n = 3a.5a =
\n = 32 . 3a.5 a = 3a
\n = 9.5a =
\n = 9.5a = 1
\n = 5a= <\/p>\n
——————\u00dcsl\u00fc Denklemler——————–<\/p>\n
1- Tabanlar\u0131 E\u015fit Olan Denklemler: <\/p>\n
KURAL:8 Tabanlar\u0131 e\u015fit olan \u00fcsl\u00fc denklemlerin \u00fcsleri de e\u015fittir.
\n a \u00b9 0, a \u00b9 -1, a \u00b9 1 olmak \u00fczere am = an Þ m=n dir
\n \u00d6RNEK\/ 1- 2x = 25 Þ x=5 tir.
\n 2- 3x = 81 Þ 3x= 34 Þ x=4 t\u00fcr.<\/p>\n
3- 2x+8 = 8 oldu\u011funa g\u00f6re, x=?
\n 2x+8 = 2x . 28 olup
\n 2x . 28 = 8 yerine konur ise, burdan 8 = 23 olup
\n 2x . 28 = 23
\n 2x = 23\u00b8 28
\n 2x = 23-8
\n 2x = 2-5 olup burdan x = -5 bulunur.<\/p>\n
\u00d6RNEK \/ e\u015fitli\u011fini sa\u011flayan x de\u011ferini bulal\u0131m.<\/p>\n
\u00c7\u00d6Z\u00dcM \/ 5x+1-(2-x) = (53)x-3
\n 5x+1-2+x= 53(x-3)
\n 52x-1= 53x-9 (Tabanlar e\u015fit olup \u00fcsler e\u015fit olmal\u0131d\u0131r.)
\n 2x-1 = 3x-9
\n 2x \u20133x = -9+1
\n -x = -8
\n x = 8<\/p>\n
2- \u00dcsleri e\u015fit olan denklemler:<\/p>\n
KURAL 8 \u00dcsleri e\u015fit olan denklemlerde \u00fcs tek say\u0131 ise tabanlar\u0131 e\u015fit, \u00fcs \u00e7ift say\u0131 ise tabanlar e\u015fit yada biri di\u011ferinin ters i\u015faretlisine e\u015fittir.
\n n tek say\u0131 ve an = bn Þ a=b dir.
\n n \u00e7ift s\u0131y\u0131 ve an = bn Þ a=b veya a = -b dir.
\n \u00d6RNEK\/ 1- x3=53Þ x=5 tir.
\n 2- (x+7)3=(3x-11)3 e\u015fitli\u011fini sa\u011flayan x de\u011ferini bulal\u0131m.<\/p>\n
\u00c7\u00f6z\u00fcm: 3=3 yani \u00fcsler e\u015fit oldu\u011fundan tabanlarda e\u015fit olmak zorundad\u0131r. Burdan,
\n (x+7) = (3x-11) olup parantezleri a\u00e7al\u0131m
\n x+7 = 3x-11
\n 7+11= 3x-x
\n 18 = 2x
\n x =
\n x = 9<\/p>\n
\u00d6RNEK \/ (2X+3)4= (X-2)4 e\u015fitli\u011fini sa\u011flayan x de\u011ferlerini bulal\u0131m.<\/p>\n
\u00c7\u00d6Z\u00dcM \/ 4\u00e7ift say\u0131 oldu\u011fu i\u00e7in
\n (2x+3)4= (X-2)4 Þ
\n 2x+3= x-2 Veya 2x+3= -(x-2)
\n 2x-x= -2-3 Veya 2x+3= -x+2
\n x=5 Veya 2x+x= 2-3
\n 3x = -1
\n x= <\/p>\n
KURAL 8 xn = 1 \u015feklinde olan denklemler.<\/p>\n
Bu t\u00fcr denklemlerin \u00e7\u00f6z\u00fcm\u00fcnde 3 durum vard\u0131r.<\/p>\n
Xn = 1 Þ<\/p>\n
\u00d6RNEK \/ 1- 18 = 1 dir. \u00c7\u00fcnk\u00fc 1 in t\u00fcm reel kuvvetleri 1 dir.
\n 2- 50 = 1 dir. \u00c7\u00fcnk\u00fc 0 d\u0131\u015f\u0131ndaki t\u00fcm reel say\u0131lar\u0131n 0 \u0131nc\u0131 kuvvetleri 1 dir.
\n 3- (-1)6 = 1 dir. \u00c7\u00fcnk\u00fc (-1) in t\u00fcm \u00e7ift kuvvetleri 1 dir.
\n 4- 53x-15 = 1 ise x=?<\/p>\n
\u00c7\u00f6z\u00fcm: 53x-15 = 1 ise
\n 3x-15 = 0 olmal\u0131d\u0131r,burdan
\n 3x = 15
\n x = 15\u00b83
\n x = <\/p>\n
\u00d6RNEK \/ (5x+3)7 = 1 ise x de\u011ferini hesaplay\u0131n.<\/p>\n
\u00c7\u00d6Z\u00dcM: (5x+3)7 = 17 (17=1 olup ) Burdan bu e\u015fitli\u011fin tabanlar\u0131 e\u015fit olmal\u0131d\u0131r.
\n (5x+3) = 1
\n 5x+3 = 1
\n 5x = 1-3
\n 5x = -2
\n x =
\n \u00d6RNEK \/ (x+3)x-2= 1 e\u015fitli\u011fini sa\u011flayan x de\u011ferini bulal\u0131m.<\/p>\n
\u00c7\u00d6Z\u00dcM \/ 1. DURUM..: x+3=1Þx=1-3
\n x=-2——(\u00aa)
\n 2. DURUM..: x-2=0–(\u00aa)
\n x=2——-(\u00aa) Bu k\u00f6k \u00fcss\u00fc s\u0131f\u0131r yapmad\u0131\u011f\u0131 i\u00e7in al\u0131n\u0131r.
\n 3. DURUM…: X+3= -1
\n x=-4——(\u00aa) Bu k\u00f6k yaz\u0131ld\u0131\u011f\u0131nda \u00fcs \u00e7ift say\u0131 olaca\u011f\u0131 i\u00e7in, bu k\u00f6k de al\u0131n\u0131r. O halde denklemi sa\u011flayan x de\u011ferleri : -4 , -2 , 2 dir.
\n \u00d6RNEK \/ i\u015fleminin sonucunu \u00fcsl\u00fc ifade olarak yazal\u0131m.<\/p>\n
\u00c7\u00d6Z\u00dcM \/ = 6.10x<\/p>\n
=3.5x<\/p>\n
=
\n =2.2x
\n =21 . 2x
\n =21+x<\/p>\n","protected":false},"excerpt":{"rendered":"
TANIM: : a bir reel ger\u00e7el say\u0131 ve n\u00ceZ+ olsun. a.a.a…a=an olacak \u015fekilde, n tane a\u2019n\u0131n \u00e7arp\u0131m\u0131 olan an e \u00fcsl\u00fc ifadeler denir. \u00d6rnek\/ a) 3.3.3.3=34 b) c) UYARI :8 a bir reel say\u0131 ve n\u00ceZ+ olmak \u00fczere a+a+a+…+a = n.a oldu\u011fu i\u00e7in an ile n.a ifadeleri birbirine kar\u0131\u015ft\u0131r\u0131lmamal\u0131d\u0131r. Yani an \u00b9 n.a d\u0131r. \u00d6rnek …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1404,1403],"tags":[7299,7300,7298],"class_list":["post-3101","post","type-post","status-publish","format-standard","hentry","category-matematik-odevleri","category-odevler","tag-bir-reel-sayinin-ussu","tag-uslu-ifadelerde-dort-islem","tag-uslu-sayilar"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3101","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=3101"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3101\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=3101"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=3101"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=3101"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}