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2.10 Breuke - optelling en aftrekking

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Wiskunde

Gewone breuke

Opvoeders afdeling

Memorandum

14. a) nommers

b) ekwivalente

c) veelvoude

d) tellers

e) nommer

f) breuke

g) onegte

h) vereenvoudig

15.2 a)

= 12 21 size 12{ { { size 8{"12"} } over { size 8{"21"} } } } {} + 14 21 size 12{ { { size 8{"14"} } over { size 8{"21"} } } } {}

= 26 21 size 12{ { { size 8{"26"} } over { size 8{"21"} } } } {}

= 1 5 21 size 12{ { { size 8{5} } over { size 8{"21"} } } } {}

b)

= 5 10 size 12{ { { size 8{5} } over { size 8{"10"} } } } {} + 6 10 size 12{ { { size 8{6} } over { size 8{"10"} } } } {}

= 11 10 size 12{ { { size 8{"11"} } over { size 8{"10"} } } } {}

= 1 1 10 size 12{ { { size 8{1} } over { size 8{"10"} } } } {}

c)

= 36 45 size 12{ { { size 8{"36"} } over { size 8{"45"} } } } {} - 25 45 size 12{ { { size 8{"25"} } over { size 8{"45"} } } } {}

= 11 45 size 12{ { { size 8{"11"} } over { size 8{"45"} } } } {}

d)

= 4 6 size 12{ { { size 8{4} } over { size 8{6} } } } {} - 3 6 size 12{ { { size 8{3} } over { size 8{6} } } } {}

= 1 6 size 12{ { { size 8{1} } over { size 8{6} } } } {}

16.

a)

= 11 2 3 size 12{"11" { { size 8{2} } over { size 8{3} } } } {} + 1 7 size 12{ { { size 8{1} } over { size 8{7} } } } {}

= 11 14 21 size 12{"11" { { size 8{"14"} } over { size 8{"21"} } } } {} + 3 21 size 12{ { { size 8{3} } over { size 8{"21"} } } } {}

p = 11 17 21 size 12{"11" { { size 8{"17"} } over { size 8{"21"} } } } {}

b)

= 3 1 4 − 1 9 size 12{3 { { size 8{1} } over { size 8{4} } } - { { size 8{1} } over { size 8{9} } } } {}

= 3 9 36 − 4 36 size 12{ { { size 8{9} } over { size 8{"36"} } } - { { size 8{4} } over { size 8{"36"} } } } {}

t = 3 5 36 size 12{ { { size 8{5} } over { size 8{"36"} } } } {}

= 6 3 4 size 12{ { { size 8{3} } over { size 8{4} } } } {} – (3 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} + 1 2 3 size 12{ { { size 8{2} } over { size 8{3} } } } {} )

= 6 3 4 size 12{ { { size 8{3} } over { size 8{4} } } } {} – 3 3 6 size 12{ { { size 8{3} } over { size 8{6} } } } {} + 4 6 size 12{ { { size 8{4} } over { size 8{6} } } } {}

= 6 3 4 size 12{ { { size 8{3} } over { size 8{4} } } } {} – 4 1 6 size 12{ { { size 8{1} } over { size 8{6} } } } {}

= 2 9 12 size 12{ { { size 8{9} } over { size 8{"12"} } } } {} - 2 12 size 12{ { { size 8{2} } over { size 8{"12"} } } } {}

g = 2 7 12 size 12{ { { size 8{7} } over { size 8{"12"} } } } {}

d)

= 9 7 8 size 12{ { { size 8{7} } over { size 8{8} } } } {} - (4 9 12 size 12{ { { size 8{9} } over { size 8{"12"} } } } {} + 8 12 size 12{ { { size 8{8} } over { size 8{"12"} } } } {} )

= 9 7 8 size 12{ { { size 8{7} } over { size 8{8} } } } {} - 5 5 12 size 12{ { { size 8{5} } over { size 8{"12"} } } } {}

= 4 7 8 size 12{ { { size 8{7} } over { size 8{8} } } } {} - 5 12 size 12{ { { size 8{5} } over { size 8{"12"} } } } {}

= 4 21 24 size 12{ { { size 8{"21"} } over { size 8{"24"} } } } {} - 10 24 size 12{ { { size 8{"10"} } over { size 8{"24"} } } } {}

v = 4 11 24 size 12{ { { size 8{"11"} } over { size 8{"24"} } } } {}

Leerders afdeling

Inhoud

Aktiwiteit: optelling en aftrekking van breuke [lu 1.7.3]

14. Die optelling en aftrekking van breuke

KOM ONS HERSIEN

Die antwoorde van die volgende vrae is hieronder versteek.

Omkring hulle soos jy hulle vind en voltooi dan die sinne.

a b h t t s o n k o m m a
v e r e e n v o u d i g h
e d b l k o a e n r a j f
e k a l e e a m d o e p b
l h s e n m l e i n r o r
v m r r s e d r g e j o e
o n t s a r f s s g h g u
u s n x l m g b t t p k
d e e k w i v a l e n t e
e l o y o n f k u v w

a) Ons kan breuke optel of aftrek slegs indien die .................................................. dieselfde is.

b) Indien die noemers verskil, moet ons .................................................. breuke met dieselfde noemers vind.

c) Ons kan die gemeenskaplike noemer maklik d.m.v. .................................................. vind.

d) Ons tel slegs die .................................................. bymekaar.

e) Die .................................................. bly onveranderd wanneer ons optel of aftrek.

f) Wanneer ons gemengde getalle optel, tel ons eers die natuurlike getalle bymekaar en dan

die ..................................................

g) Wanneer ons gemengde getalle aftrek, kan ons dit eers na ................................................. breuke verander.

h) Antwoorde moet sover moontlik altyd .................................................. word.

15.1 Onthou jy nog?

Dus: 1 3 + 4 5 5 15 + 12 15 17 15 1 2 15 alignl { stack { size 12{ { { size 8{1} } over { size 8{3} } } + { { size 8{4} } over { size 8{5} } } } {} #= { { size 8{5} } over { size 8{"15"} } } + { { size 8{"12"} } over { size 8{"15"} } } {} # = { { size 8{"17"} } over { size 8{"15"} } } {} #=1 { { size 8{2} } over { size 8{"15"} } } {} } } {}

15.2 Bereken die volgende:

a) x = 4 7 + 2 3 size 12{x= { { size 8{4} } over { size 8{7} } } + { { size 8{2} } over { size 8{3} } } } {}

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b) y = 1 2 + 3 5 size 12{y= { { size 8{1} } over { size 8{2} } } + { { size 8{3} } over { size 8{5} } } } {}

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c) d = 4 5 − 5 9 size 12{d= { { size 8{4} } over { size 8{5} } } - { { size 8{5} } over { size 8{9} } } } {}

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d) k = 2 3 − 1 2 size 12{k= { { size 8{2} } over { size 8{3} } } - { { size 8{1} } over { size 8{2} } } } {}

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16. Werk saam met ’n maat en bereken:

a) p = 7 2 3 + 4 1 7 size 12{p=7 { { size 8{2} } over { size 8{3} } } +4 { { size 8{1} } over { size 8{7} } } } {}

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b) t = 5 1 4 − 2 1 9 size 12{t=5 { { size 8{1} } over { size 8{4} } } - 2 { { size 8{1} } over { size 8{9} } } } {}

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c) g = 6 3 4 − 2 1 2 + 1 2 3 size 12{g=6 { { size 8{3} } over { size 8{4} } } - left (2 { { size 8{1} } over { size 8{2} } } +1 { { size 8{2} } over { size 8{3} } } right )} {}

___________________________________________________

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d) v = 9 7 8 − 3 3 4 + 1 2 3 size 12{v=9 { { size 8{7} } over { size 8{8} } } - left (3 { { size 8{3} } over { size 8{4} } } +1 { { size 8{2} } over { size 8{3} } } right )} {}

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17. UITDAGING!

Deel in groepe van drie. Voltooi die volgende tabel deur die aantal ure wat julle verlede week aan tuiswerk bestee het, in te vul:

NAAM Maan Dins Woens Don Vry
Bv. Nomsa 1 1 2 size 12{1 { { size 8{1} } over { size 8{2} } } } {} 2 1 4 size 12{2 { { size 8{1} } over { size 8{4} } } } {} 3 3 4 size 12{3 { { size 8{3} } over { size 8{4} } } } {} 1 1 2 size 12{1 { { size 8{1} } over { size 8{2} } } } {} 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {}
1. ............................................... ............ ............ ............ ............ ............
2. ............................................... ............ ............ ............ ............ ............
3. ............................................... ............ ............ ............ ............ ............

Beantwoord die volgende vrae:

a) Hoeveel uur het elkeen van jou groeplede altesaam aan tuiswerk bestee?

1. _________________________________

2. _________________________________

3. _________________________________

b) Who Wie het die meeste tyd aan huiswerk bestee? _______________________

c) Wie het die minste geleer? _________________________________

d) Bereken die verskil tussen jou antwoorde by b en c.

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

e) Vra nou ’n ander groep om jul antwoorde te kontroleer.

Assessering

Leeruitkomste 1: Die 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.

Assesseringstandaard 1.7: Dit is duidelik wanneer die leerder skat en bereken deur geskikte bewerkings vir probleme wat die volgende behels, kies en gebruik:

1.7.3: optelling, aftrekking en vermenigvuldiging van gewone breuke.

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Source:  OpenStax, Wiskunde graad 7. OpenStax CNX. Oct 21, 2009 Download for free at http://cnx.org/content/col11076/1.2
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