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5. 7x 3 xy ÷ 3x + 6 5x 2 y ÷ 5x 10 3x 2 12 size 12{ { {7x} over {3 ital "xy"} } div { {3x+6} over {5x rSup { size 8{2} } y} } div { {5x - "10"} over {3x rSup { size 8{2} } - "12"} } } {}

6. 5x 2 + 5x x 2 x 5x + 5 x 2 1 size 12{ { { { {5x rSup { size 8{2} } +5x} over {x rSup { size 8{2} } - x} } } over { { {5x+5} over {x rSup { size 8{2} } - 1} } } } } {} (Hier is ‘n breuk gedeel deur ‘n breuk – herskryf dit eers soos in 4)

C. Optelling van breuke

  • Jy weet alreeds dat die optelling en aftrekking van breuke heelwat moeiliker is as vermenigvuldiging en deling. Dit is omdat ons slegs gelyksoortige breuke (met eenderse noemers) kan optel en aftrek. As die noemers verskil, moet jy die kleinste gemene veelvoud (KGV) van die noemers soek en dan elke breuk oor hierdie noemer skryf. Vereenvoudig dan die breuk. Vereenvoudig weer die antwoord so ver moontlik. Hier volg voorbeelde – al die stappe word getoon:

Vereenvoudig:

1. 5 abx 2 cx + 4 ac 3x + cx 2a size 12{ { {5 ital "abx"} over {2 ital "cx"} } + { {4 ital "ac"} over {3x} } + { { ital "cx"} over {2a} } } {} (KGV = 6acx)

5 abx 2 cx × 3a 3a + 4 ac 3x × 2 ac 2 ac + cx 2a × 3 cx 3 cx size 12{ left ( { {5 ital "abx"} over {2 ital "cx"} } times { {3a} over {3a} } right )+ left ( { {4 ital "ac"} over {3x} } times { {2 ital "ac"} over {2 ital "ac"} } right )+ left ( { { ital "cx"} over {2a} } times { {3 ital "cx"} over {3 ital "cx"} } right )} {} = 15 a 2 bx 6 acx + 8a 2 c 2 6 acx + 3c 2 x 2 6 acx size 12{ { {"15"a rSup { size 8{2} } ital "bx"} over {6 ital "acx"} } + { {8a rSup { size 8{2} } c rSup { size 8{2} } } over {6 ital "acx"} } + { {3c rSup { size 8{2} } x rSup { size 8{2} } } over {6 ital "acx"} } } {} = 15 a 2 bx + 8a 2 c 2 + 3c 2 x 2 6 acx size 12{ { {"15"a rSup { size 8{2} } ital "bx"+8a rSup { size 8{2} } c rSup { size 8{2} } +3c rSup { size 8{2} } x rSup { size 8{2} } } over {6 ital "acx"} } } {}

2. a + b 2 + b + c 3 a + c 6 size 12{ { {a+b} over {2} } + { {b+c} over {3} } - { {a+c} over {6} } } {} (LCD = 6) Let baie fyn op hoe die tekens hieronder hanteer word!

3 a + b 6 + 2 b + c 6 a + c 6 size 12{ { {3 left (a+b right )} over {6} } + { {2 left (b+c right )} over {6} } - { {a+c} over {6} } } {} = 3 a + b + 2 b + c a + c 6 size 12{ { {3 left (a+b right )+2 left (b+c right ) - left (a+c right )} over {6} } } {} = 3a + 3b + 2b + 2c a c 6 size 12{ { {3a+3b+2b+2c - a - c} over {6} } } {} = 2a + 5b + c 6 size 12{ { {2a+5b+c} over {6} } } {}

3. a + 3 a 2 4 + 1 3a + 6 + 2 5a 10 size 12{ { {a+3} over {a rSup { size 8{2} } - 4} } + { {1} over {3a+6} } + { {2} over {5a - "10"} } } {}

Om die Kleinste Gemene Noemer te vind, faktoriseer eers die noemers!

a + 3 a + 2 a 2 + 1 3 a + 2 + 2 5 a 2 size 12{ { {a+3} over { left (a+2 right ) left (a - 2 right )} } + { {1} over {3 left (a+2 right )} } + { {2} over {5 left (a - 2 right )} } } {}

Kan jy sien die KGV is 3×5×(a+2)(a–2)?

a + 3 a + 2 a 2 × 15 15 + 1 3 a + 2 × 5 a 2 5 a 2 + 2 5 a 2 × 3 a + 2 3 a + 2 size 12{ left ( { {a+3} over { left (a+2 right ) left (a - 2 right )} } times { {"15"} over {"15"} } right )+ left ( { {1} over {3 left (a+2 right )} } times { {5 left (a - 2 right )} over {5 left (a - 2 right )} } right )+ left ( { {2} over {5 left (a - 2 right )} } times { {3 left (a+2 right )} over {3 left (a+2 right )} } right )} {}

= 15 a + 3 + 5 a 2 + 6 a + 2 15 a + 2 a 2 size 12{ { {"15" left (a+3 right )+5 left (a - 2 right )+6 left (a+2 right )} over {"15" left (a+2 right ) left (a - 2 right )} } } {} = 15 a + 45 + 5a 10 + 6a + 12 15 a + 2 a 2 size 12{ { {"15"a+"45"+5a - "10"+6a+"12"} over {"15" left (a+2 right ) left (a - 2 right )} } } {} = 26 a + 47 15 a + 2 a 2 size 12{ { {"26"a+"47"} over {"15" left (a+2 right ) left (a - 2 right )} } } {}

Oefening:

Vereenvoudig die volgende uitdrukkings deur van faktorisering gebruik te maak:

1. a x 2 a x + 5a 2x size 12{ { {a} over {x rSup { size 8{2} } } } - { {a} over {x} } + { {5a} over {2x} } } {}

2. 1 3 + 2x + 1 2x x 1 3x size 12{ { {1} over {3} } + { {2x+1} over {2x} } - { {x - 1} over {3x} } } {}

3. 4a 4b 2a 2 2b 2 3 2a 2b size 12{ { {4a - 4b} over {2a rSup { size 8{2} } - 2b rSup { size 8{2} } } } - { {3} over {2a - 2b} } } {}

4. 1 2 a 2 + 2 3 a + 1 3 4 a 3 size 12{ { {1} over {2} } left (a - 2 right )+ { {2} over {3} } left (a+1 right ) - { {3} over {4} } left (a - 3 right )} {}

  • ‘n Laaste wenk. Ons sou 2 x 1 3 x + 3 × 9 x + 3 1 x size 12{ { {2 left (x - 1 right )} over {3 left (x+3 right )} } times { {9 left (x+3 right )} over { left (1 - x right )} } } {} beter kon vereenvoudig as (1–x) so gelyk het: (x–1).
  • Ons kan hierdie verandering maak as ons die teken van die hele tweeterm ook verander: (1–x) = –(x–1) omdat –(x–1) = –x + 1, en dit is 1–x. Voltooi self die probleem.

Assessering

Leeruitkomstes(LUs)
LU 1
Getalle, Bewerkings en VerwantskappeDie leerder is in staat om getalle en die verwantskappe daarvan te herken, te beskryf en voor te stel, en om tydens probleemoplossing bevoeg en met selfvertroue te tel, te skat, te bereken en te kontroleer.
Assesseringstandaarde(ASe)
Ons weet dit as die leerder:
1.1 die historiese ontwikkeling van getallestelsels in ’n verskeidenheid historiese en kulturele kontekste (insluitend plaaslik) beskryf en illustreer;
1.2 rasionale getalle (insluitend baie klein getalle in wetenskaplike notasie) herken, gebruik en voorstel, en sonder huiwering tussen ekwivalente vorms in gepaste kontekste beweeg;
1.3 probleme in konteks oplos, insluitend kontekste wat gebruik word om ‘n bewustheid van ander leerareas, asook van menseregte-, sosiale, ekonomiese en omgewingsake, te bevorder, soos:
1.3.1 finansiële kontekste (insluitend wins en verlies, begrotings, rekeninge, lenings, enkelvoudige en saamgestelde rente, huurkoop, wisselkoerse, kommissie, huur en die bankwese);
1.3.2 metings in die Natuurwetenskappe en Tegnologie;
1.4 probleme oplos wat verhouding, koers en proporsie (direkte en omgekeerde) behels;
1.5 skat en bereken deur geskikte bewerkings vir probleme te kies en te gebruik en die redelikheid van resultate te beoordeel (insluitend meetprobleme wat rasionale benaderings van irrasionale getalle behels);
1.6 ’n verskeidenheid van tegnieke en instrumente (insluitend tegnologie) gebruik om berekeninge doeltreffend en met die nodige mate van akkuraatheid te doen, insluitend die volgende reëls en betekenisse van eksponente (leerders behoort in staat te wees om hierdie reëls en betekenisse slegs in berekeninge te gebruik):
1.6.1 x n × x m = x n + m
1.6.2 x n  x m = x n – m
1.6.3 x 0 = 1
1.6.4 x –n = 1 x n size 12{ { {1} over {x rSup { size 8{n} } } } } {}
1.7 die eienskappe van rasionale getalle herken, beskryf en gebruik.
LU 2
Patrone, Funksies en AlgebraDie leerder is in staat om patrone en verwantskappe te herken, te beskryf en voor te stel, en probleme op te los deur algebraïese taal en vaardighede te gebruik.
Ons weet dit as die leerder:
2.1 op verskillende maniere ‘n verskeidenheid numeriese en meetkundige patrone en verwantskappe ondersoek deur dit voor te stel en te veralgemeen, en deur die reëls onderliggend daaraan te verduidelik en te bewys (insluitend patrone in natuurlike en kulturele vorms en patrone wat die leerders self geskep het);
2.7 die distributiewe wet en manipuleringsvaardighede wat in graad 8 ontwikkel is gebruik om die volgende te doen:
  • bepaal die produk van tweeterme;
  • faktoriseer algebraïse uitdrukkings (beperk tot gemene faktore en die verskil van vierkante);
2.8 die eksponentwette gebruik om uitdrukkings te vereenvoudig en vergelykings op te los;
2.9 faktorisering om algebraïese uitdrukkings te vereenvoudig en vergelykings op te los gebruik.

Memorandum

Bespreking

Terminologie

  • Leerders gebruik dikwels metodes vir gebruik met uitdrukkings wanneer hulle met vergelykings werk (byvoorbeeld, hulle kan noemers in uitdrukkings weglaat), en andersom. Hou ‘n ogie hierop en leer hulle om die konteks te ondersoek voor hulle blindelings voortgaan.
  • Vermenigvuldiging en faktorisering werk omgekeerd – die leerders moet hiervan bewus word. Dit maak in elk geval die werk makliker om te bemeester.
  • As leerders nie terme en faktore kan onderskei nie, sal hulle nie uitdrukkings behoorlik kan manipuleer nie. As dit nodig blyk, kan hulle meer oefeninge gegee word.
  • In die algemeen vind leerders breuke moeilik. Dit is dalk goed om in sulke gevalle met nie–algebraïse breuke te begin om die basis te vestig.

TOETS

1. Vereenvoudig die volgende uitdrukkings deur gelyksoortige terme bymekaar te maak.

1.1 3 a 2 + 3 a 2 – 6 a + 3 a – 4 + 1

1.2 2 y 2 – 1 y + 2 y 2 – 6 + 2 y – 9

1.3 8 x 2 – (5 x + 12 x 2 – 1) + x – 4

1.4 (3 a a 2 ) – [(2 a 2 – 11) – (5 a – 3)]

2. Gee die antwoorde tot die volgende probleme in die eenvoudigste vorm:

2.1 Tel 3 x 2 + 5 x – 1 by x 2 – 3 x

2.2 Bereken die som van 2 a + 3 b – 5 en 3 + 2 b – 7 a

2.3 Trek 6 a + 7 af van 5 a 2 + 2 a + 2

2.4 Hoeveel is 3 a – 8 b + 3 minder as a + b + 2?

3. Vereenvoudig deur vermenigvuldiging en laat antwoord in eenvoudigste vorm:

3.1 (3 x 2 ) × (2 x 3 )

3.2 ( abc ) ( a 2 c ) (2 b 2 )

3.3 abc ( a 2 c + 2 b 2 )

3.4 –3 a (2 a 2 – 5 a )

3.5 ( a – 2 b ) ( a + 2 b )

3.6 (3 – x 2 ) (2 x 2 + 5)

3.7 ( x – 5 y ) 2

3.8 (2 – b ) (3 a + c )

Memorandum

1.1 6 a 2 – 3 a – 3

1.2 4 y 2 + y – 15

1.3 – 4 x 2 – 4 x – 3

1.4 – 3 a 2 + 8 a + 8

2.1 4 x 2 + 2 x – 1

2.2 – 5 a + 5 b – 2

2.3 5 a 2 – 4 a – 5

2.4 – 2 a + 9 b – 1

3.1 6 x 5

3.2 2 a 3 b 3 c 2

3.3 a 3 bc 2 + 2 ab 3 c

3.4 – 6 a 3 + 15 a 2

3.5 a 2 – 4 b 2

3.6 –2 x 4 + x 2 + 15

3.7 x 2 – 10 xy + 25 y 2

3.8 6 a + 2 c – 3 ab bc

TOETS

1. Bepaal die Grootste Gemene Faktor van hierdie drie uitdrukkings: 6 a 2 c 2 en 2 ac 2 en 10 ab 2 c 3 .

2. Faktoriseer die volgende uitdrukkings volledig deur gemene faktore te bepaal:

2.1 12 a 3 + 3 a 4

2.2 –5 xy – 15 x 2 y 2 – 20 y

2.3 6 a 2 c 2 – 2 ac 2 + 10 ab 2 c 3

3. Faktoriseer hierdie verskille van kwadrate volledig:

3.1 a 2 – 4

3.2 1 9 a 2 9b 2 size 12{ { {1} over {9} } a rSup { size 8{2} } - 9b rSup { size 8{2} } } {}

3.3 x 4 – 16 y 4

3.4 1 – a 4 b 4

4. Faktoriseer hierdie uitdrukkings so ver as moontlik:

4.1 3 x 2 – 27

4.2 2 a – 8 ab 2

4.3 a 2 – 5 a – 6

4.4 a 2 + 7 a + 6

5. Vereenvoudig die volgende breuke deur van faktorisering gebruik te maak:

5.1 3a 2 3 6a + 6 size 12{ { {3a rSup { size 8{2} } - 3} over {6a+6} } } {}

5.2 6x 2 y 6y 2x 2 size 12{ { {6x rSup { size 8{2} } y - 6y} over {2x - 2} } } {}

5.3 a 2 9 2 × 1 4a 2 12 a size 12{ { {a rSup { size 8{2} } - 9} over {2} } times { {1} over {4a rSup { size 8{2} } - "12"a} } } {}

5.4 3x + 6 5 ÷ x 2 4 15 size 12{ { {3x+6} over {5} } div { {x rSup { size 8{2} } - 4} over {"15"} } } {}

5.5 abx 2 cx + 2 ac 3x + 3 cx 2a size 12{ { { ital "abx"} over {2 ital "cx"} } + { {2 ital "ac"} over {3x} } + { {3 ital "cx"} over {2a} } } {}

5.6 2a x 2 3a x + a 2x size 12{ { {2a} over {x rSup { size 8{2} } } } - { {3a} over {x} } + { {a} over {2x} } } {}

5.7 4a 4b 2a 2 2b 2 3 2a 2b size 12{ { {4a - 4b} over {2a rSup { size 8{2} } - 2b rSup { size 8{2} } } } - { {3} over {2a - 2b} } } {}

5.8 2 3 a + 2 + 1 3 a 1 1 4 a 5 size 12{ { {2} over {3} } left (a+2 right )+ { {1} over {3} } left (a - 1 right ) - { {1} over {4} } left (a - 5 right )} {}

Memorandum

1. 2 ac 2

2.1 3 a 3 (4 + a 2 )

2.2 –5 y ( x + 3 x 2 y + 4)

2.3 2 ac 2 (3 a – 1 + 5 b 2 c )

3.1 ( a + 2) ( a – 2)

3.2 1 3 a + 3b 1 3 a 3b size 12{ left ( { {1} over {3} } a+3b right ) left ( { {1} over {3} } a - 3b right )} {}

3.3 ( x 2 + 4 y 2 ) ( x + 2 y ) ( x – 2 y )

3.4 (1 + a 2 b 2 ) (1 + ab ) (1 – ab )

4.1 3 ( x + 3) ( x – 3)

4.2 2 a (1 + 4 b ) (1 – 4 b )

4.3 ( a + 1) ( a – 6)

4.4 ( a + 1) ( a + 6)

5.1 a 1 2 size 12{ { {a - 1} over {2} } } {}

5.2 3 y ( x + 1)

5.3 a + 3 8a size 12{ { {a+3} over {8a} } } {}

5.4 9 x 2 size 12{ { {9} over {x - 2} } } {}

5.5 3a 2 bx + 4a 2 c 2 + 9c 2 x 2 6 acx size 12{ { {3a rSup { size 8{2} } ital "bx"+4a rSup { size 8{2} } c rSup { size 8{2} } +9c rSup { size 8{2} } x rSup { size 8{2} } } over {6 ital "acx"} } } {}

5.6 4a 5 ax 2x 2 size 12{ { {4a - 5 ital "ax"} over {2x rSup { size 8{2} } } } } {}

5.7 a 7b 2 a + b a b size 12{ { {a - 7b} over {2 left (a+b right ) left (a - b right )} } } {}

5.8 3a + 9 4 size 12{ { {3a+9} over {4} } } {}

Questions & Answers

what does preconceived mean
sammie Reply
physiological Psychology
Nwosu Reply
How can I develope my cognitive domain
Amanyire Reply
why is communication effective
Dakolo Reply
Communication is effective because it allows individuals to share ideas, thoughts, and information with others.
effective communication can lead to improved outcomes in various settings, including personal relationships, business environments, and educational settings. By communicating effectively, individuals can negotiate effectively, solve problems collaboratively, and work towards common goals.
it starts up serve and return practice/assessments.it helps find voice talking therapy also assessments through relaxed conversation.
miss
Every time someone flushes a toilet in the apartment building, the person begins to jumb back automatically after hearing the flush, before the water temperature changes. Identify the types of learning, if it is classical conditioning identify the NS, UCS, CS and CR. If it is operant conditioning, identify the type of consequence positive reinforcement, negative reinforcement or punishment
Wekolamo Reply
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Wekolamo
because it helps many people around the world to understand how to interact with other people and understand them well, for example at work (job).
Manix Reply
Agreed 👍 There are many parts of our brains and behaviors, we really need to get to know. Blessings for everyone and happy Sunday!
ARC
A child is a member of community not society elucidate ?
JESSY Reply
Isn't practices worldwide, be it psychology, be it science. isn't much just a false belief of control over something the mind cannot truly comprehend?
Simon Reply
compare and contrast skinner's perspective on personality development on freud
namakula Reply
Skinner skipped the whole unconscious phenomenon and rather emphasized on classical conditioning
war
explain how nature and nurture affect the development and later the productivity of an individual.
Amesalu Reply
nature is an hereditary factor while nurture is an environmental factor which constitute an individual personality. so if an individual's parent has a deviant behavior and was also brought up in an deviant environment, observation of the behavior and the inborn trait we make the individual deviant.
Samuel
I am taking this course because I am hoping that I could somehow learn more about my chosen field of interest and due to the fact that being a PsyD really ignites my passion as an individual the more I hope to learn about developing and literally explore the complexity of my critical thinking skills
Zyryn Reply
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Jonathan
and having a good philosophy of the world is like a sandwich and a peanut butter 👍
Jonathan
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Kelu Reply
interpersonal relationships
Abdulfatai Reply
What would be the best educational aid(s) for gifted kids/savants?
Heidi Reply
treat them normal, if they want help then give them. that will make everyone happy
Saurabh
What are the treatment for autism?
Magret Reply
hello. autism is a umbrella term. autistic kids have different disorder overlapping. for example. a kid may show symptoms of ADHD and also learning disabilities. before treatment please make sure the kid doesn't have physical disabilities like hearing..vision..speech problem. sometimes these
Jharna
continue.. sometimes due to these physical problems..the diagnosis may be misdiagnosed. treatment for autism. well it depends on the severity. since autistic kids have problems in communicating and adopting to the environment.. it's best to expose the child in situations where the child
Jharna
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Jharna
results you'll get.. please consult a therapist to know what suits best on your child. and last as a parent. I know sometimes it's overwhelming to guide a special kid. but trust the process and be strong and patient as a parent.
Jharna
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Source:  OpenStax, Wiskunde graad 9. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col11055/1.1
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