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1. (a + b) (a – b) = a 2 – ab + ab – b 2 = a 2 – b 2 (vereenvoudig)

2. (a + 2) (a + 3) = a 2 + 3a + 2a + 6 = a 2 + 5a + 6

3. (a + b) (a + b) = a×a +ab + ba + b×b = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2 (vereenvoudig)

4. (a + b) (c + d) = ac + ad + bc + bd (hierdie antwoord kan nie vereenvoudig word nie)

  • Die antwoord op die soort probleem in vraag 1 hierbo het die vorm van ‘n verskil van vierkante.
  • Die antwoorde op 2 en 3 is drieterme. Ons gaan nou probeer om drieterme te faktoriseer.
  • Die eerste feit om te onthou is dat nie alle drieterme gefaktoriseer kan word nie .

Werk agteruit deur probleem 2:

a 2 + 5a + 6 = a 2 + 3a + 2a + 6 = (a + 2) (a + 3).

  • So is dit duidelik waar die a 2 vandaan kom, en die 5a en die 6.

Faktoriseer nou a 2 + 7a + 12 = ( ) ( ) deur twee geskikte tweeterme in die twee paar hakies te skryf.

  • As jy die tweeterme in die hakies uitvermenigvuldig soos jy in aktiwiteit 2.2 geleer is, kan jy jou antwoord toets. Hou aan en toets telkens jou antwoorde tot jy seker is hoe om dit te doen. Doen dieselfde in die volgende oefeninge:
  • Elke drieterm het ‘n maat in die tweede kolom; soek hulle uit:

A. a 2 – 5a – 6 1. (x + 2)(x + 3)

B. a 2 – a – 6 2. (x – 2)(x + 3)

C. a 2 – 5a + 6 3. (x + 1)(x – 6)

D. a 2 + 7a + 6 4. (x – 2)(x – 3)

E. a 2 + 5a + 6 5. (x + 1)(x + 6)

F. a 2 + 5a – 6 6. (x – 1)(x + 6)

G. a 2 + a – 6 7. (x + 2)(x – 3)

H. a 2 – 7a + 6 8. (x – 1)(x – 6)

  • Faktoriseer nou die volgende drieterme op dieselfde manier. Die laaste twee is moeiliker as die eerste vier!
  1. a 2 + 3a + 2
  2. a 2 + a – 12
  3. a 2 – 4a + 3
  4. a 2 – 9a + 20
  5. a 2 + ab – 12b 2
  6. 2a 2 – 18a + 40

Aktiwiteit 4

Om faktorisering te gebruik in die vereenvoudiging van breuke, en in die optelling, vermenigvuldiging en deling van breuke

[lu 1.2, 1.6, 2.9]

A. Vereenvoudiging van algebraïese breuke

Twee van die volgende vier breuke kan vereenvoudig word, en twee nie. Watter twee kan?

2 + a 2 a size 12{ { {2+a} over {2 - a} } } {}

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

4 + x x + 4 size 12{ { {4+x} over {x+4} } } {}

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

Jy het seker nou al agtergekom dat dit baie moeite is om te faktoriseer. Hoekom doen ons dit?

  • Hierdie breuk kan nie vereenvoudig word soos dit staan nie: 6a 2 b 6b 2a 2 size 12{ { {6a rSup { size 8{2} } b - 6b} over {2a - 2} } } {} . Dis omdat ons nie terme mag kanselleer nie. As ons die som uitdrukkings na produk uitdrukkings kan verander (deur faktorisering) sal ons die faktore kan kanselleer, en sodoende klaar kan vereenvoudig.

6a 2 b – 6b = 6b(a 2 – 1) = 6b (a + 1) (a – 1) en 2a – 2 = 2(a – 1)

  • Dus is die motivering vir faktorisering die behoefte aan vereenvoudiging.

Dus: 6a 2 b 6b 2a 2 size 12{ { {6a rSup { size 8{2} } b - 6b} over {2a - 2} } } {} = 6b a + 1 a 1 2 a 1 size 12{ { {6b left (a+1 right ) left (a - 1 right )} over {2 left (a - 1 right )} } } {} = 3b a + 1 1 size 12{ { {3b left (a+1 right )} over {1} } } {} = 3b(a + 1) .

Dit is baie belangrik om volledig te faktoriseer.

Oefening:

Faktoriseer beide teller en noemer, en vereenvoudig:

1 12 a + 6b 2a + b size 12{ { {"12"a+6b} over {2a+b} } } {}

2 x 2 9 x + 3 size 12{ { {x rSup { size 8{2} } - 9} over {x+3} } } {}

3 2 a + 1 a 1 6 a + 1 2 size 12{ { {2 left (a+1 right ) left (a - 1 right )} over {6 left (a+1 right ) rSup { size 8{2} } } } } {}

4 5a 2 5 5a + 5 size 12{ { {5a rSup { size 8{2} } - 5} over {5a+5} } } {}

B. Vermenigvuldiging en deling van breuke

  • Die gewone reëls om breuke te vermenigvuldig en te deel bly steeds van toepassing. Bestudeer die volgende voorbeelde – let veral op na die faktorisering en kansellering.

4x 3 y 6y 2 ÷ xy 3x 2 × 2 xy 2 3x size 12{ { {4x rSup { size 8{3} } y} over {6y rSup { size 8{2} } } } div { { ital "xy"} over {3x rSup { size 8{2} } } } times { {2 ital "xy" rSup { size 8{2} } } over {3x} } } {} = 4x 3 y 6y 2 × 3x 2 xy × 2 xy 2 3x size 12{ { {4x rSup { size 8{3} } y} over {6y rSup { size 8{2} } } } times { {3x rSup { size 8{2} } } over { ital "xy"} } times { {2 ital "xy" rSup { size 8{2} } } over {3x} } } {} = 4x 4 3 size 12{ { {4x rSup { size 8{4} } } over {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} } } {} = a + 3 a 3 2 × 1 4a a 3 size 12{ { { left (a+3 right ) left (a - 3 right )} over {2} } times { {1} over {4a left (a - 3 right )} } } {} = a + 3 8a size 12{ { { left (a+3 right )} over {8a} } } {}

3a + 6 5 ÷ a 2 4 10 size 12{ { {3a+6} over {5} } div { {a rSup { size 8{2} } - 4} over {"10"} } } {} = 3a + 6 5 × 10 a 2 4 size 12{ { {3a+6} over {5} } times { {"10"} over {a rSup { size 8{2} } - 4} } } {} = 3 a + 2 5 × 10 a + 2 a 2 size 12{ { {3 left (a+2 right )} over {5} } times { {"10"} over { left (a+2 right ) left (a - 2 right )} } } {} = 6 a 2 size 12{ { {6} over {a - 2} } } {}

Oefening:

Vereenvoudig:

1. 2 ab 2 b 3 c × 9 ac 2 4b ÷ 3 ac 2b 2 size 12{ { {2 ital "ab" rSup { size 8{2} } } over {b rSup { size 8{3} } c} } times { {9 ital "ac" rSup { size 8{2} } } over {4b} } div { {3 ital "ac"} over {2b rSup { size 8{2} } } } } {}

2. 2 a + 1 a 2 2 a 2 3 a + 3 × 9 a + 1 a + 3 2 4 a 2 ÷ 3 a + 1 a + 3 2 a 2 2 size 12{ { {2 left (a+1 right ) left (a - 2 right ) rSup { size 8{2} } } over { left (a - 2 right ) rSup { size 8{3} } left (a+3 right )} } times { {9 left (a+1 right ) left (a+3 right ) rSup { size 8{2} } } over {4 left (a - 2 right )} } div { {3 left (a+1 right ) left (a+3 right )} over {2 left (a - 2 right ) rSup { size 8{2} } } } } {}

3. 4a 2 + 8a 2b + 4 × 3 b 2 + 2 3a 2 + 6a size 12{ { {4a rSup { size 8{2} } +8a} over {2b+4} } times { {3 left (b rSup { size 8{2} } +2 right )} over {3a rSup { size 8{2} } +6a} } } {}

4. x 2 1 5x 5 ÷ x + 1 2 15 x + 15 size 12{ { {x rSup { size 8{2} } - 1} over {5x - 5} } div { { left (x+1 right ) rSup { size 8{2} } } over {"15"x+"15"} } } {}

Questions & Answers

so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
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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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