zad 1 zapisz w postaci jednej potegi
a) (6³)^4*(6² )³ =
b)9*3^4*27/3^15 :3^5 =
c)1/32*(-1/16)*(1/2)^4 =
d)3^5+3^5+3^5/3³ =
zad 2 oblicz
a) 2^12:(-10^7:5^7) =
b)(4 1/2)³ *(1 1/2)-³ =
c)(-3)^7*(-6)^7/18^8 =
d) 4-² *4^7*4^-8 =
zad 3 oblicz zapisując wynik w notacji wykladniczje
a) 3,3*10^ -3*2*10^ -6 =
b)2,6*10^7*5*10^8 =
c)2*10^7/5*10^ -5
d)7*10^ -5/14*10^ -9
zad 4 Usuń niewymierność z mianownika
a)2/√5
b)3/2√3 prosiłbym o rozwiązanie tych zadań daje naj :)
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Zad. 1. a) (6³)⁴ ·(6²)³ = 6¹² · 6⁶ = 6¹⁸
b) (9·3⁴·27)/3¹⁵ : 3⁵ = (3²·3⁴·3³)/3¹⁵ : 3⁵ = 3⁹/3¹⁵ : 3⁵ = 3⁻⁶ : 3⁵ = 3⁻¹¹
(mam nadzieję, że dobrze zrozumiałam Twój zapis zad. b) )
c)1/32·(-1/16)·(½)⁴ = -(½)⁵·(½)⁴·(½)⁴= - (½)¹³
d) (3⁵+3⁵+3⁵)/3³ = (3·3⁵)/3³ = 3⁶/3³ = 3³
Zad. 2. a) 2¹²/(-10⁷:5⁷) = 2¹²/(-2⁷) = -2⁵ = -32
b) (4½)³ ·(1½)⁻³ = (9/2)³·(3/2)⁻³ = (9/2)³·(⅔)³ = (9/2·⅔)³ = 3³ =27
c) ( (-3)⁷·(-6)⁷)/18⁸ = (3⁷·6⁷)/18⁸ = 18⁷/18⁸= 1/18
d) 4⁻²·4⁷·4⁻⁸ = 4⁻²⁺⁷⁻⁸ = 4⁻³ = (³ = 1/64
Zad. 3. a) 3,3·10⁻³ ·2·10⁻⁶ = 6,6⁻⁹
b) 2,6·10⁷·5·10⁸ = 13·10¹⁵ = 1,3·10¹⁶
c) (2·10⁷)/(5·10⁻⁹) = 0,4·10¹² = 4·10¹¹
d) (7·10⁻⁵)/(14·10⁻⁹) = 0,5 ·10⁴ = 5·10³
Zad. 4. a) 2/√5 = (2√5)/(√5·√5) = (2√5)/5
b) 3/(2√3) = (3√3)/(2√3·√3) = (3√3)/6 = √3/2