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Vereenvoudiging van breuke

In sommige gevalle van die vereenvoudiging van 'n algebraïese uitdrukking, sal die uitdrukking 'n breuk wees. Byvoorbeeld,

x 2 + 3 x x + 3

het 'n kwadraat in die teller en 'n binomiaal (kwadratiese tweeterm) in die noemer. Jy kan die verskillende metodes van faktorisering gebruik om die uitdrukking te vereenvoudig.

x 2 + 3 x x + 3 = x ( x + 3 ) x + 3 = x solank x - 3

As x 3 is, sal die noemer, x - 3 , 0 wees en die breuk ongedefinieer.

Vereenvoudig: 2 x - b + x - a b a x 2 - a b x

  1. Gebruik groepering vir die teller en uithaal van 'n gemene faktor vir die noemer in hierdie voorbeeld.

    = ( a x - a b ) + ( x - b ) a x 2 - a b x = a ( x - b ) + ( x - b ) a x ( x - b ) = ( x - b ) ( a + 1 ) a x ( x - b )
  2. Die vereenvoudigde antwoord is:

    = a + 1 a x

Vereenvoudig: x 2 - x - 2 x 2 - 4 ÷ x 2 + x x 2 + 2 x

  1. = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) ÷ x ( x + 1 ) x ( x + 2 )
  2. = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) × x ( x + 2 ) x ( x + 1 )
  3. Die vereenvoudigde antwoord is

    = 1

Vereenvoudiging van breuke

  1. Vereenvoudig:
    (a) 3 a 15 (b) 2 a + 10 4
    (c) 5 a + 20 a + 4 (d) a 2 - 4 a a - 4
    (e) 3 a 2 - 9 a 2 a - 6 (f) 9 a + 27 9 a + 18
    (g) 6 a b + 2 a 2 b (h) 16 x 2 y - 8 x y 12 x - 6
    (i) 4 x y p - 8 x p 12 x y (j) 3 a + 9 14 ÷ 7 a + 21 a + 3
    (k) a 2 - 5 a 2 a + 10 ÷ 3 a + 15 4 a (l) 3 x p + 4 p 8 p ÷ 12 p 2 3 x + 4
    (m) 16 2 x p + 4 x ÷ 6 x 2 + 8 x 12 (n) 24 a - 8 12 ÷ 9 a - 3 6
    (o) a 2 + 2 a 5 ÷ 2 a + 4 20 (p) p 2 + p q 7 p ÷ 8 p + 8 q 21 q
    (q) 5 a b - 15 b 4 a - 12 ÷ 6 b 2 a + b (r) f 2 a - f a 2 f - a
  2. Vereenvoudig: x 2 - 1 3 × 1 x - 1 - 1 2

Optel en aftrek van breuke

Deur gebruik te maak van die konsepte wat ons geleer het in die vereenvoudiging van breuke, kan ons nou eenvoudige breuke optel en aftrek. Om breuke op te tel of af te trek, moet ons daarop let dat ons slegs breuke kan optel of aftrek wat dieselfde noemer het. Dus moet ons eers al die noemers herlei na dieselfde noemer en dan die bewerkings van optelling of aftrekking doen. Dit word genoem die vind van die kleinste gemeenskaplike noemer of veelvoud.

Byvoorbeeld, om 1 2 en 3 5 op te tel, let ons op dat die kleinste gemene noemer 10 is. Dus moet ons die eerste breuk se noemer vermenigvuldig met 5 en die tweede breuk met 2 om beide se noemers te herlei na breuke met dieselfde noemer. Dit gee: 5 10 en 6 10 . Nou kan ons die breuke optel. As ons dit doen, kry ons 11 10 .

Vereenvoudig die volgende uitdrukking: x - 2 x 2 - 4 + x 2 x - 2 - x 3 + x - 4 x 2 - 4

  1. x - 2 ( x + 2 ) ( x - 2 ) + x 2 x - 2 - x 3 + x - 4 ( x + 2 ) ( x - 2 )
  2. Ons maak al die noemers dieselfde sodat ons die breuke kan optel of aftrek. Die kleinste gemeenskaplike noemer is ( x - 2 ) ( x + 2 ) .

    x - 2 ( x + 2 ) ( x - 2 ) + ( x 2 ) ( x + 2 ) ( x + 2 ) ( x - 2 ) - x 3 + x - 4 ( x + 2 ) ( x - 2 )
  3. Aangesien al die breuke dieselfde noemer het, kan ons hulle almal skryf as een breuk met die toepaslike bewerkingstekens.

    x - 2 + ( x 2 ) ( x + 2 ) - x 3 + x - 4 ( x + 2 ) ( x - 2 )
  4. 2 x 2 + 2 x - 6 ( x + 2 ) ( x - 2 )
  5. 2 ( x 2 + x - 3 ) ( x + 2 ) ( x - 2 )

Twee interessante wiskundige bewyse

Ons kan die konsepte wat ons in hierdie hoofstuk geleer het, gebruik om twee interessante wiskundige bewyse te illustreer. Die eerste bewering is dat n 2 + n ewe is vir alle n Z . Die tweede is 'n bewys dat n 3 - n deelbaar is deur 6 vir alle n Z . Voor ons kan toon dat hierdie twee bewerings waar is, moet ons eers kennis neem van sekere ander wiskundige reëls.

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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 10) [caps]. OpenStax CNX. Aug 04, 2011 Download for free at http://cnx.org/content/col11328/1.4
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