SlideShare a Scribd company logo
1 of 6
PROJECT ICT
MAGISTER PENDIDIKAN MATEMATIKA PPS
UNSRI
Oleh :

RAHMAWATI
0602268131016
Pangkat tak
sebenarnya
Bilangan berpangkat
adalah bilangan
perkalian berulang
an = a x a x a x a…x a sampai n factor
jika a bilangan bulat, m dan n bilangan asli, berlaku
a3 x a2 = ( a x a x a ) x ( a x a )
= ( a x a x a x a x a)
= ( a5)
a3 x a2 = a3 + 2 = a5
jadi an x am = an + m
jika a bilangan bulat tak nol, m dan n bilangan asli
dengan m > n maka
a6 : a3 =( a x a x a x a x a x a )
( a x a x a)
=(axaxa)
= ( a 3)
a6 : a3 = ( a 6 – 3 ) = ( a 3)
Jadi, am : an = am – n



jika a adalah bilangan bulat, m dan n merupakan
bilangan asli maka
(a3)2 = ( a3 ) x ( a3 )
= ( a x a x a ) x ( a x a x a)
= a6
(a3)2 = a3x2 = a6
Jadi (am)n = amxn = amn


jika a, b bilangan bulat dan n bilangan asli, berlaku:
( a x b )3 = (a x b) x (a x b) x(a x b)
=axaxaxbxbxb
= a3 x b3
( a x b )3 = a3 x b3
Jadi ( a x b )m = am x bm

More Related Content

Viewers also liked

Viewers also liked (8)

Ema fil2013
Ema fil2013Ema fil2013
Ema fil2013
 
Ftth and-cable-glands-revolutionised
Ftth and-cable-glands-revolutionisedFtth and-cable-glands-revolutionised
Ftth and-cable-glands-revolutionised
 
Industrial switchgear
Industrial switchgearIndustrial switchgear
Industrial switchgear
 
Magnetic Message
Magnetic MessageMagnetic Message
Magnetic Message
 
Villagio qatar slideshow
Villagio qatar slideshowVillagio qatar slideshow
Villagio qatar slideshow
 
Pax Goguryeoana
Pax GoguryeoanaPax Goguryeoana
Pax Goguryeoana
 
Heteronyms
HeteronymsHeteronyms
Heteronyms
 
Cabinet Members, Philippines
Cabinet Members, PhilippinesCabinet Members, Philippines
Cabinet Members, Philippines
 

PROJECT Presentation rahmawati

  • 1. PROJECT ICT MAGISTER PENDIDIKAN MATEMATIKA PPS UNSRI Oleh : RAHMAWATI 0602268131016
  • 3. an = a x a x a x a…x a sampai n factor jika a bilangan bulat, m dan n bilangan asli, berlaku a3 x a2 = ( a x a x a ) x ( a x a ) = ( a x a x a x a x a) = ( a5) a3 x a2 = a3 + 2 = a5 jadi an x am = an + m
  • 4. jika a bilangan bulat tak nol, m dan n bilangan asli dengan m > n maka a6 : a3 =( a x a x a x a x a x a ) ( a x a x a) =(axaxa) = ( a 3) a6 : a3 = ( a 6 – 3 ) = ( a 3) Jadi, am : an = am – n 
  • 5.  jika a adalah bilangan bulat, m dan n merupakan bilangan asli maka (a3)2 = ( a3 ) x ( a3 ) = ( a x a x a ) x ( a x a x a) = a6 (a3)2 = a3x2 = a6 Jadi (am)n = amxn = amn
  • 6.  jika a, b bilangan bulat dan n bilangan asli, berlaku: ( a x b )3 = (a x b) x (a x b) x(a x b) =axaxaxbxbxb = a3 x b3 ( a x b )3 = a3 x b3 Jadi ( a x b )m = am x bm