Aula 5 - Apostilas do curso de pré cálculo PRIMA-FURG

a \u2208 n \u2208 N a n = b a n b \u2022 a n = a.a.....a | {z } n f atores \u2022 a 1 = a \u2022 a 0 = 1 \u2022 a \u2212 n = 1 a n \u2022 1 n = 1 \u2200 n 99K ( par a todo n ) \u2022 0 n = 0 \u2200 n > 0 99K ( par a todo n maior do q ue 0) \u2022 10 n = 1 ..., n N × 10 n 136 , 5=1 , 365 × 10 2 0 , 000847 = 8 , 47 × 10 \u2212 4 \u2022 x 2 n > 0 , \u2200 x \u2208 \u2022 3 × 3 × 3 × 3 = 3 4 = 81 \ue012 2 3 \ue013 · \ue012 2 3 \ue013 = \ue012 2 3 \ue013 2 = 4 9 2 2 3 4 9 0 , 5 · 0 , 5 · 0 , 5 = 0 , 5 3 = 0 , 125 0 , 5 0 , 125 5 · 5 · 5 = 5 3 2 · 2 · 2 · 2 · 2=2 5 3 4 · 3 4 · 3 4 · = \ue012 3 4 \ue013 3 2 4 = 2 · 2 · 2 · 2 6 3 = 6 · 6 · 6 · 6 · 6 · 6 \ue012 2 5 \ue013 2 = 2 5 · 2 5 3 4 = 3 = 3 · 3 · 3 · 3 = 81 \ue012 2 5 \ue013 = \ue012 2 5 \ue013 = 2 5 · 2 5 = 4 25 (0 , 6) 3 = (0 , 6) 3 = (0 , 6) · (0 , 6) · (0 , 6) = 0 , 216 2 5 = 2 · 2 · 2 · 2 · 2 = 32 12 2 = 12 · 12 = 144 6 1 = 6 3 0 = 1 2 5 = 2 . 2 . 2 . 2 . 2 = 32 5 2 = 5 . 5 = 25 5 0 = 1 120 1 = 120 ( 2 3 ) \u2212 1 = 3 2 0 100 = 0 ( \u2212 2) 3 = ( \u2212 2)( \u2212 2)( \u2212 2) = \u2212 8 ( \u2212 2) 4 = ( \u2212 2)( \u2212 2)( \u2212 2)( \u2212 2) = 16 ( \u2212 2) \u2212 1 = \u2212 1 2 10 7 = 10 . 000 . 000 ( \u2212 6) 2 = ( \u2212 6)( \u2212 6) = 36 6 \u2212 2 = \ue000 1 6 \ue001 2 = 1 36 1 4 = 1 ( 3 7 ) \u2212 2 = ( 7 3 ) 2 = 49 9 10 \u2212 4 = 0 , 0001 (0 , 2) 3 = 0 , 008 268 . 000 = 2 , 68 × 10 0 , 00023 = 2 , 3 × 10 \u2212 4 2 . 500 . 000 . 000 = 2 , 5 × 10 m, n \u2208 Z a, b \u2208 \u2022 a m .a n = a m + n \u2022 a m ÷ a n = a m a n = a m \u2212 n \u2022 ( a m ) n = a m.n \u2022 ( a.b ) m = a m .b m \u2022 ( a b ) m = a m b m 10 3 · 10 5 = 10 3+5 = 10 8 6 4 · 6=6 4+1 = 6 5 \ue012 2 5 \ue013 4 · \ue012 2 5 \ue013 2 = \ue012 2 5 \ue013 4+2 = \ue012 2 5 \ue013 6 (0 , 1) 4 · (0 , 1) 2 · (0 , 1) = (0 , 1) 4+2+1 = (0 , 1) 7 10 5 : 10 3 = 10 5 \u2212 3 = 10 2 6 4 : 6 = 6 4 \u2212 1 = 6 3 \ue012 3 5 \ue013 6 : \ue012 3 5 \ue013 3 = \ue012 3 5 \ue013 6 \u2212 3 = \ue012 3 5 \ue013 3 (0 , 7) 8 : (0 , 7) 5 = (0 , 7) 8 \u2212 5 = (0 , 7) 3 (10 5 ) 3 = 10 5 · 3 = 10 15 (2 6 ) 2 = 2 6 · 2 = 2 12 [(2 5 ) 2 ] 4 = 2 5 · 2 · 4 = 2 40 " \ue012 1 2 \ue013 4 # 6 = \ue012 1 2 \ue013 4 · 6 = \ue012 1 2 \ue013 24 (3 · 7) 2 = 3 2 · 7 2 (2 · 5) 4 = 2 4 · 5 4 (3 · 4 · 9) 3 = 3 3 · 4 3 · 9 3 (6 : 4) 5 = 6 5 : 4 5 (10 : 3) 4 = 10 4 : 3 4 \ue012 3 5 \ue013 4 = 3 5 2 2 . 2 \u2212 3 = 1 / 2 99K V 2 2 . 2 \u2212 3 = 2 \u2212 1 = 1 2 (3 2 ) 3 = 3 5 99K F (3 2 ) 3 = 3 2 · 3 = 3 10 x +1 = 10 x . 10 99K V 10 x . 10 = 10 x . 10 = 10 x +1 a 3 2 = ( a 3 ) 2 99K F a 3 2 = a 9 5 5 x = 5 1+ x 99K F 5 5 x = 5 ÷ 5 x = 5 1 \u2212 x 2 n : 2 n \u2212 1 = 2 99K V 2 n : 2 n \u2212 1 = 2 n \u2212 ( n \u2212 1) = 2 \ue001 n \u22126 n +1 = 2 = 2 5 x +1 . 5 x \u2212 1 = 5 2 x 99K V 5 x +1 . 5 x \u2212 1 = 5 x + 6 1+ x \u22126 1 = 5 2 x 10 2 10 5 = 10 3 99K F 10 2 10 5 = 10 2 \u2212 5 = 10 \u2212 3 ( a x +2 ) 3 = a 3 x +6 99K V ( a x +2 ) 3 = a ( x +2) · 3 = a 3 x +6 (2 x ) x = 2 2 x 99K F (2 x ) x = 2 x 10 x +2 : 10 x +1 = 10 99K V 10 x +2 : 10 x +1 = 10 x +2 \u2212 ( x +1) = 10 \ue001 x +2 \u2212 \ue001 x \u2212 1 = 10 2 \u2212 1 = 10 = 10 ( a \u2212 2 .b 3 ) \u2212 2 = a 4 .b \u2212 6 99K V ( a \u2212 2 .b 3 ) \u2212 2 = a ( \u2212 2) . ( \u2212 2) .b 3 . ( \u2212 2) = a 4 .b \u2212 6 a, b \u2208 n \u2208 N \u2217 , n \u221a b = a a b n \u221a . n \u221a b = a \u21d4 a n = b 2 2 \u221a b = \u221a b 1 1 1 1 \u221a b = b \u2022 b \u2265 0 n a \u221a 25 = 5 \u21d0 \u21d2 5 2 = 25 3 \u221a 27 = 3 \u21d0 \u21d2 3 3 = 27 3 \u221a 0=0 \u21d0 \u21d2 0 3 = 0 5 q 1 32 = 1 2 \u21d0 \u21d2 \ue000 1 2 \ue001 5 = 1 32 \u2022 b 0 n a \u221a 4 / \u2208 R \u2212 4 4 \u221a \u2212 16 / \u2208 R 6 \u221a \u2212 100 / \u2208 R \u2022 b 0 n a 5 \u221a \u2212 32 = \u2212 2 \u21d0 \u21d2 ( \u2212 2) 5 = \u2212 32 7 \u221a \u2212 1 = \u2212 1 \u21d0 \u21d2 ( \u2212 1) 7 = \u2212 1 3 \u221a \u2212 8 = \u2212 2 \u21d0 \u21d2 ( \u2212 2) 3 = \u2212 8 3 q \u2212 1 1000 = \u2212 1 10 \u21d0 \u21d2 ( \u2212 1 10 ) 3 = \u2212 1 1000 ( n \u221a b ) n = b n \u221a a · n \u221a b = n \u221a a · b n \u221a a n \u221a b = n p a b , b 6 = 0 n p m \u221a a = n · m \u221a a ( n \u221a a ) p = n \u221a a p , p \u2208 N \u2217 n \u221a a m = n · p \u221a a m · p n ÷ p \u221a a m ÷ p , p \u2208 N \u2217 ( \u221a 4) 2 = 2 2 = 4 \u221a 8 · \u221a 2 = \u221a 8 · 2 = \u221a 16 = 4 3 \u221a 5 3 \u221a 7 = 3 q 5 7 p \u221a 5 = 4 \u221a 5 ( 3 \u221a 9) 2 = 3 \u221a 9 2 = 3 \u221a 81 12 \u221a 10 6 12 \u221a 10 6 6 12 \u221a 10 6 = 12 ÷ 6 \u221a 10 6 ÷ 6 = \u221a 10 6 \u221a a 3 3 6 \u221a a 3 = 6 ÷ 3 \u221a a 3 ÷ 3 = 2 \u221a a 1 = \u221a a 6 \u221a 27 b 3 6 \u221a 27 b 3 = 6 \u221a 3 3 b 3 = 6 p (3 · b ) 3 = 6 ÷ 3 p (3 · b ) 3 ÷ 3 = \u221a 3 b 5 \u221a a 10 = 5 ÷ 5 \u221a a 10 ÷ 5 = 1 \u221a a 2 = a 2 8 \u221a a 6 · b 2 = 8 p ( a 3 b ) 2 = 8 ÷ 2 p ( a 3 b ) 2 ÷ 2 = 4 \u221a a 3 b 10 \u221a 32 = 10 \u221a 2 5 = 10 ÷ 5 \u221a 2 5 ÷ 5 = \u221a 2 3 \u221a a 6 · x. 3 \u221a a 6 · x = 3 \u221a a 6 · 3 \u221a x 3 ÷ 3 \u221a a 6 ÷ 3 · 3 \u221a x = a 2 3 \u221a x \u221a a 2 b = \u221a a 2 · \u221a b = a \u221a b 3 \u221a a 6 · x = 3 \u221a a 6 · 3 \u221a x = 3 ÷ 3 \u221a a 6 ÷ 3 · 3 \u221a x = a 2 3 \u221a x 3 \u221a a 6 b 3 x = 3 \u221a a 6 · 3 \u221a b 3 · 3 \u221a x = a 2 b 3 \u221a x 10 \u221a 2048 = 10 \u221a 2 11 = 10 \u221a 2 10 · 2 = 10 \u221a 2 10 · 10 \u221a 2=2 10 \u221a 2 \u221a a 2 + b 2 6 = a + b \u221a a 2 \u2212 b 2 6 = a \u2212 b \ue01b ( a 2 + b 2 ) ( a 2 \u2212 b 2 ) ( a + b ) ( a \u2212 b ) p \u2208 Z , n \u2208 N \u2217 a \u2208 \u2217 + \u21d2 a p n = n \u221a a p a = 0 \u21d2 ( 0 p n = 0 p n > 0 0 p n p n 0 a \u2208 \u2217 \u2212 \u21d2 ( a p n a p n = n \u221a a p ( \u2212 4) 1 2 = \u221a \u2212 4 / \u2208 R a 5 3 = 3 \u221a a 5 = 3 \u221a a 3 · a 2 = a 3 \u221a a 2 ( \u2212 16) 3 4 = 4 p ( \u2212 16) 3 = 4 \u221a \u2212 4096 / \u2208 R 2 1 2 = \u221a 2 ( x 2 · y 4 ) 1 4 = 4 p x 2 · y 4 = y · 4 \u221a x 2 = y \u221a x m \u2212 3 4 = 1 4 \u221a m 3 ( \u2212 27) 1 3 = 3 \u221a \u2212 27 = \u2212 3 \u221a 7=7 1 2 6 \u221a a 5 = a 5 6 3 \u221a x 2 = x 2 3 1 3 \u221a 7 2 = 1 7 2 3 = 7 \u2212 2 3 99K 3 5 \u221a a 3 = 3 · a \u2212 3 5 99K \u221a a 2 b = a \u221a b a \u221a b = \u221a a 2 b p x 4 y = x 2 \u221a y x 2 \u221a y = p x 4 · y 2 \u221a 7 3 \u221a a 2 n \u221a a m = n · p \u221a a m · p 2 \u221a 7 = 2 · 3 \u221a 7 1 · 3 = 6 \u221a 7 3 3 \u221a a 2 = 3 · 2 \u221a a 2 · 2 = 6 \u221a a 4 + \u221a a 5 \u221a a 2 99K \u221a a 5 \u221a a 2 (2 , 5) = 10 10 \u221a a 1 · 5 , 10 \u221a a 2 · 2 10 \u221a a 5 , 10 \u221a a 4 3 \u221a a, \u221a ab, 4 \u221a 2 a 3 3 \u221a a \u221a ab 4 \u221a 2 a (3 , 2 , 4) = 12 12 \u221a a 4 , 12 p ( ab ) 6 , 12 p (2 a 3 ) 3 12 \u221a a 4 , 12 \u221a a 6 b 6 , 12 \u221a 8 a 9 2 \u221a a + 5 \u221a a \u2212 2 \u221a a 2 \u221a a + 5 \u221a a \u2212 3 \u221a a = (2 + 5 \u2212 3) \u221a a = 4 \u221a a \u221a 2+7 \u221a 2 \u2212 3 \u221a 2+2 \u221a 2 = (1 + 7 \u2212 3 + 2) \u221a 2 = 7 \u221a 2 2 3 \u221a 5 + 3 \u221a 5 \u2212 6 3 \u221a 5 = (2 + 1 \u2212 6) 3 \u221a 5 = \u2212 3 3 \u221a 5 3 \u221a 2 + \u221a 5 \u2212 \u221a 2+4 \u221a 5 3 \u221a 2 + \u221a 5 \u2212 \u221a 2+4 \u221a 5 = (3 \u2212 1) \u221a 2 + (1 + 4) \u221a 5 = 2 \u221a 2+5 \u221a 5 3 \u221a 50 \u2212 2 \u221a 18 + \u221a 98 50 = 2 · 5 2 18 = 2 · 3 2 98 = 2 · 7 2 3 \u221a 50 \u2212 2 \u221a 18 + \u221a 98 = 3 \u221a 2 · 5 2 \u2212 2 \u221a 2 · 3 2 + \u221a 2 · 7 2 = 3 · 5 \u221a 2 \u2212 2 · 3 \u221a 2+7 \u221a 2 = 15 \u221a 2 \u2212 6 \u221a 2+7 \u221a 2 = (15 \u2212 6 + 7) \u221a 2 = 16 \u221a 2 \u221a 12 \u2212 \u221a 50 \u2212 \u221a 3+6 \u221a 2 12 = 2 2 · 3 50 = 5 2 · 2 \u221a 12 \u2212 \u221a 50 \u2212 \u221a 3+6 \u221a 2 = \u221a 2 2 · 3 \u2212 \u221a 5 2 · 2 \u2212 \u221a 3+6 \u221a 2=2 \u221a 3 \u2212 5 \u221a 2 \u2212 \u221a 3+6 \u221a 2 = \u221a 3 + \u221a 2 n \u221a a · n \u221a b = n \u221a a · b \u221a 3 · \u221a 5 · \u221a 6 = \u221a 3 · 5 · 6 = \u221a 90 = \u221a 3 2 · 2 · 5=3 \u221a 10 3 \u221a a · 3 \u221a a 5 = 3 \u221a 3 · a 5 = 3 \u221a a 6 = a 2 \ue000 \u221a 5 + \u221a 3 \ue001 · \ue000 \u221a 5 \u2212 \u221a 3 \ue001 ( a + b ) · ( a \u2212 b ) = a \u2212 b 2 \ue000 \u221a 5 \ue001 2 \u2212 \ue000 \u221a 3 \ue001 2 = 5 \u2212 3=2 3 \u221a x + 3 · 3 \u221a x \u2212 2 = 3 p ( x + 3)( x \u2212 2) = 3 \u221a x 2 \u2212 2 x + 3 x \u2212 6 = 3 \u221a x 2 + x \u2212 6 3 \u221a 3 · \u221a 3 = 6 \u221a 3 2 · 6 \u221a 3 3 = 6 \u221a 3 5 6 \u221a a · 3 \u221a a · \u221a a = 6 \u221a a · 6 \u221a a 2 · 6 \u221a a 3 = 6 \u221a a · a 2 · a 3 = 6 \u221a a 6 = a \u221a ab · 6 \u221a a 3 b 2 = 6 \u221a a 3 b 3 · 6 \u221a a 3 b 2 = 6 \u221a a 3 · b 3 · a 3 · b 2 = 6 \u221a a 6 b 5 = a 6 \u221a b 5

Aula 5 - Apostilas do curso de pré cálculo PRIMA-FURG

Cálculo I

FURG

Cálculo I

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