Umumiy o‘rta ta’lim maktablarining 7- sinfi uchun darslik
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(121—122):
121. 1) + + + + 4 4 4 4 4; 3) + + ; c c c 2) + + + 6 6 6 6; 4) + + + + . a a a a a M a s h q l a r 56 122. 1) + + 2 2 2 ; m m m 5) + + + 14 4244 3 21 marta 3 3 ... 3; 2) + + 17 17 17 ; ab ab ab 6) + + + 14 4244 3 17 marta 5 5 ... 5; 3) - + - ( 2 ) ( 2 ); c d c d 7) + + + 1442443 marta ... ; n m m m 4) - + - + - (3 ) (3 ) (3 ); b a b a b a 8) + + + 144244 3 marta ... . k b b b Ko‘paytmani daraja shaklida yozing (123—125): 123. 1) × × × × 2 2 2 2 2; 2) × × × × 1 1 1 1 1 ; 3 3 3 3 3 3) × × 3 3 3 4 4 4 ; 4) - × - × - × - ( 2,7) ( 2,7) ( 2,7) ( 2,7). 124. 1) × × × × ; x x x x x 3) × × (2 ) (2 ) (2 ); a a a 2) × × × × ; m m m m m 4) - × - × - × - ( 3 ) ( 3 ) ( 3 ) ( 3 ). b b b b 125. 1) - × - × - ( ) ( ) ( ); x y x y x y 3) × 3 3 2 2 ; x x 2) + × + ( ) ( ); a b a b 4) × × × × . m m m m m n n n n n Ko‘paytmaning daraja shaklidagi yozuvidan foydalanib, ifodani soddalashtiring (126 — 128): 126. 1) 2·2·2·15; 3) 5·5·8·8·8·2·2; 2) 4·4·4·4·21; 4) 6·6·7·7·3·3·3. 127. 1) 1,2·1,2·2·2·5·5; 2) 0,5·0,5·0,5·2·2·4·4; 3) × × × × × 1 1 1 1 7 7 7 7 0,3 0,3 ; 4) × × × × 2 2 2 3 3 3 2,3 2,3. 128. 1) 9·9·9· a·a·a ; 3) - × - × × ( ) ( ) ; x x x y y y x y x y 2) x·x·x·x ·3·3; 4) - × - × - × × (8 ) (8 ) (8 ) . a a b b a b a b a b Ifodani soddalashtiring (129–130): 129. 1) p · p · p · p + q ·q ; 3) a · a + a · a + a · a ; 2) a · a + b · b · b · b ; 4) x · x · x + x · x · x . 57 130. 1) × + × + + × 144424443 marta ... ; k c c c c c c 3) × × × + × × × 14243 14243 marta marta ... ... ; n m a a a b b b 2) × × + × × + + × × 1444442444443 marta ... ; n a a a a a a a a a 4) × × × + × × × 14243 14243 marta 17 marta 5 5 ... 5 ... . k a a a 131. Ifodani o‘qing, darajaning asosini, daraja ko‘rsatkichini ayting: 1) 3 2 ; 3) - 41 2 9 ; 5) + 15 (4 ) ; m n 2) 3 3 8 1 ; 4) - 39 ( 1,2) ; 6) 7 2 3 . a b Hisoblang (132 — 139) : 132. 1) 2 3 ; 2) 3 2 ; 3) 4 4 ; 4) 5 3 . 133. 1) 1 5 ; 2) ( - 1) 7 ; 3) 0 15 ; 4) 0 5 . 134. 1) 3 2 3 ; 2) 2 3 5 ; 3) 2 2 7 1 ; 4) 3 1 3 2 . 135. 1) (2,5) 2 ; 2) (1,7) 2 ; 3) ( - 0,2) 3 ; 4) ( - 0,2) 4 . 136. 1) ( - 5) 3 ; 2) - 5 3 ; 3) - 2 1 4 ; 2 4) - 2 1 4 . 2 137. 1) - 4 5 ( 0, 2) (0,1) ; 2) - 3 4 (0,3) ( 0,1) ; 3) 2 2 (3, 2) (1,6) ; 4) 2 2 (2, 6) (1,3) . 138. 1) 2 · ( - 3) 2 ; 2) - 5 · ( - 2) 3 ; 3) - × - 2 1 2 ( 4) ; 4) - × - 2 2 3 ( 3) . 139. 1) - × - 2 3 5 ( 5) ; 2) - × - 3 2 3 ( 3) ; 3) - - × 2 3 ( 3) 2 ; 4) - - × - 2 3 ( 3) ( 2) . 140. - - - 2 2 3 ; ( ) ; ( ) x x x ifodaning qiymatini = - 1 2 1 ; 5 x da toping. 58 141. x 2 ifodaning qiymatini x ning jadvalda keltirilgan qiymatlari uchun hisoblang: 0 1 - 1 2 - 2 3 - 3 4 - 4 5 - 5 6 - 6 142. x 3 ifodaning qiymatini x ning jadvalda ko‘rsatilgan qiy- matlari uchun hisoblang: 0 1 - 1 2 - 2 3 - 3 4 - 4 5 - 5 6 - 6 143. Quyidagi da’volarning qaysi biri to‘g‘ri, qaysi biri no- to‘g‘ri? Sababini tushuntiring. Da’vo noto‘g‘ri deb ayt- sangiz, uni rad etuvchi misol toping. 1) ikkita sonning kvadratlari teng bo‘lsa, bu sonlarning o‘zlari ham teng; 2) ikkita sonning kublari teng bo‘lsa, bu sonlarning o‘z- lari ham teng; 3) agar manfiy songa uning kvadrati qo‘shilsa, musbat son hosil bo‘ladi; 4) agar manfiy sondan uning kvadrati ayirilsa, manfiy son hosil bo‘ladi; 5) agar musbat sondan uning kvadrati ayirilsa, musbat son hosil bo‘ladi. Quyidagi da’volarning qaysi biri to‘g‘ri, qaysi biri no to‘g‘- ri? Sababini tushuntiring. Mos misollar tuzing (144–145): 144. 1) natural sonning kvadrati ixtiyoriy raqam bilan tugashi mumkin; 2) natural sonning kubi ixtiyoriy raqam bilan tugashi mumkin. 145. 1) natural sonning to‘rtinchi darajasi faqat 0; 1; 5; 6 ra- qamlaridan biri bilan tugashi mumkin. 2) natural sonning beshinchi darajasi shu son qaysi ra- qam bilan tugagan bo‘lsa, o‘sha raqam bilan tugaydi. x x 2 x x 3 59 10- Natural ko‘rsatkichli darajaning xossalari Darajaga ko‘tarish bir nechta muhim xossalarga ega. 1 - x o s s a . + × = . m n m n a a a Bir xil asosli darajalarni ko‘paytirishda asos o‘zgarmasdan qoladi, daraja ko‘rsatkichlari esa qo‘shiladi. Natural ko‘rsatkichli darajaning ta’rifiga ko‘ra × = × × × × = 1424 3 2 3 2 marta 3 marta 2 2 (2 2) (2 2 2) × = × × × × ´ × × × × = 1442443 1442443 marta marta ( ... ) ( ... ) m n m n a a a a a a a a a a ko‘paytirishning guruhlash qonuniga ko‘ra = × × × × = 14243 5 marta 2 2 2 2 2 + = × × × × = 144244 3 ( ) marta ... m n a a a a natural ko‘rsatkichli darajaning ta’rifiga ko‘ra = 2 5 . = a m + n . Shunday qilib, 2 2 · 2 3 = 2 2+3 . a m · a n = a m + n . 2- xossa. - = > ¹ : , , 0 m n m n a a a m n a . Bir xil asosli darajalarni bo‘lishda asos o‘zgarmasdan qoladi, daraja ko‘rsatkichlari esa ayiriladi . Shartga ko‘ra 5 > 3. m > n , a ¹ 0. Darajaning birinchi xossasiga ko‘ra 2 5 – 3 · 2 3 = 2 5 . a m – n · a n = a m . Shuning uchun 2 5 – 3 = 2 5 : 2 3 . a m – n = a m : a n . 60 Shunday qilib, 2 5 : 2 3 = 2 5 – 3 . a m : a n = a m – n , m > n , a ¹ 0. = ¹ 1, 0 n n a a a ekanligini ta’kidlaymiz. 3- xossa. = ( ) m n mn a a . Darajani darajaga ko‘tarishda asos o‘zgarmasdan qoladi, daraja ko‘rsatkichlar esa o‘zaro ko‘paytiriladi. Natural ko‘rsatkichli darajaning ta’rifiga ko‘ra = × = 3 2 3 3 (2 ) 2 2 = × × × × = 144424443 marta ( ) ... m n m m m m n a a a a a darajaning birinchi xossasiga ko‘ra = 2 3 + 3 = + + + = = 64748 ... marta m m m n a ko‘paytirishning ta’rifiga ko‘ra = 2 3 · 2 . = a mn . Shunday qilib, (2 3 ) 2 = 2 3 · 2 . ( a m ) n = a mn . 4- xossa. = ( ) n n n ab a b . Ko‘paytmani darajaga ko‘tarishda har bir ko‘paytuvchi shu darajaga ko‘tariladi. × = × × × × × = 144424443 3 3 marta (2 3) (2 3) (2 3) (2 3) = = 1442443 marta ( ) ( )( )...( ) n n ab ab ab ab ko‘paytirishning guruhlash va o‘rin almashtirish qonuniga ko‘ra = × × × × × = 1424 3 1424 3 3 marta 3 marta (2 2 2) (3 3 3) = × × × × × × = 14243 14243 marta marta ( ... )( ... ) n n a a a b b b natural ko‘rsatkichli darajaning ta’rifiga ko‘ra = 2 3 · 3 3 . = a n · b n . Shunday qilib, (2·3) 3 = 2 3 · 3 3 . ( ab ) n = a n b n . 61 5- xossa. = ¹ ; 0 n n n a a b b b . Kasrni darajaga ko‘tarishda uning surat va maxraji xuddi shu darajaga ko‘tariladi. Natural ko‘rsatkichli darajaning ta’rifiga ko‘ra = × × = 14243 3 2 2 2 2 3 3 3 3 3 marta = × = 14243 marta ... n a a a a b b b b n kasrlarni ko‘paytirish qoidasiga ko‘ra × × = = × × 6474 8 123 3 marta 3 marta 2 2 2 3 3 3 × × = = × × 64748 14243 marta n marta ... ... n a a a b b b natural ko‘rsatkichli darajaning ta’rifiga ko‘ra = 3 3 2 3 . = n n a b . Shunday qilib, = 3 3 3 2 2 3 3 . = ¹ , 0 n n n a a b b b . 1- masala. Hisoblang: × × × × 7 3 4 6 4 11 7 3 11 7 3 . - - × × × × = × × = × = 7 3 4 7 6 3 1 6 4 11 7 3 11 7 3 11 7 1 11 49 539. 2- masala. Yorug‘likning tarqalish tezligi 3·10 8 m/s ga ya- qin, Yerdan Quyoshgacha bo‘lgan o‘rtacha masofa 1,5 · 10 11 m. Yorug‘lik nuri Quyoshdan Yergacha bo‘lgan masofani qancha vaqtda bosib o‘tadi? Tekis harakatda bosib o‘tilgan yo‘lning s = vt formulasiga asosan: 1,5 · 10 11 = 3 ·10 8 · t , bu yerdan × × = = × = 11 3 8 1,5 10 3 10 0,5 10 500 t (s). Javob: 500 s = 8 min 20 s. · : ( ) ( ) · m n mn n n n n n n m n m n m n m n a a b b a a a a a a a a ab a b + - = = = = = |