Getal & Ruimte (12e editie) - vwo wiskunde B

'Wortels vereenvoudigen'.

2 vwo 5.3 Wortels herleiden

Wortels vereenvoudigen (5)

opgave 1

Herleid.

2p

a

\(\sqrt{80} + \sqrt{500}\)

Optellen (5)
0085 - Wortels vereenvoudigen - basis - 0ms

a

\(\sqrt{80} + \sqrt{500} = \sqrt{16} ⋅ \sqrt{5} + \sqrt{100} ⋅ \sqrt{5} = 4 \sqrt{5} + 10 \sqrt{5} \text{.}\)

1p

\(4 \sqrt{5} + 10 \sqrt{5} = 14 \sqrt{5} \text{.}\)

1p

1p

b

\(\sqrt{125}\)

FactorVoorWortelteken (1)
0086 - Wortels vereenvoudigen - basis - 0ms

b

\(\sqrt{125} = \sqrt{25} ⋅ \sqrt{5} = 5 \sqrt{5} \text{.}\)

1p

1p

c

\(-5 \sqrt{63}\)

FactorVoorWortelteken (2)
0087 - Wortels vereenvoudigen - basis - 0ms

c

\(-5 \sqrt{63} = -5 ⋅ \sqrt{9} ⋅ \sqrt{7} = -5 ⋅ 3 ⋅ \sqrt{7} = -15 \sqrt{7} \text{.}\)

1p

2p

d

\(2 \sqrt{12} - 5 \sqrt{300}\)

Optellen (6)
0088 - Wortels vereenvoudigen - basis - 0ms

d

\(2 \sqrt{12} - 5 \sqrt{300} = 2 ⋅ \sqrt{4} ⋅ \sqrt{3} - 5 ⋅ \sqrt{100} ⋅ \sqrt{3} \text{.}\)

1p

\(2 ⋅ 2 ⋅ \sqrt{3} - 5 ⋅ 10 ⋅ \sqrt{3} = 4 \sqrt{3} - 50 \sqrt{3} = -46 \sqrt{3} \text{.}\)

1p

opgave 2

Herleid.

1p

\(\sqrt{2\frac{1}{4}}\)

BreukInWortel (1)
008b - Wortels vereenvoudigen - basis - 72ms

\(\sqrt{2\frac{1}{4}} = \sqrt{\frac{9}{4}} = {\sqrt{9} \over \sqrt{4}} = \frac{3}{2} = 1\frac{1}{2} \text{.}\)

1p

3 vwo 5.5 Wortels herleiden

Wortels vereenvoudigen (6)

opgave 1

Herleid.

1p

a

\({5 \over 9 \sqrt{7}}\)

WortelInNoemer
0089 - Wortels vereenvoudigen - basis - 0ms

a

\({5 \over 9 \sqrt{7}} = {5 \over 9 \sqrt{7}} ⋅ {\sqrt{7} \over \sqrt{7}} = {5 \sqrt{7} \over 9 ⋅ 7} = \frac{5}{63} \sqrt{7} \text{.}\)

1p

1p

b

\(\sqrt{\frac{76}{81}}\)

BreukInWortel (2)
008c - Wortels vereenvoudigen - basis - 1ms

b

\(\sqrt{\frac{76}{81}} = {\sqrt{76} \over \sqrt{81}} = {\sqrt{76} \over 9} = \frac{1}{9} \sqrt{76} = \frac{1}{9} ⋅ 2 ⋅ \sqrt{19} = \frac{2}{9} \sqrt{19} \text{.}\)

1p

1p

c

\(\sqrt{2\frac{15}{17}}\)

BreukInWortel (3)
008d - Wortels vereenvoudigen - basis - 0ms

c

\(\sqrt{2\frac{15}{17}} = \sqrt{\frac{49}{17}} = {\sqrt{49} \over \sqrt{17}} = {7 \over \sqrt{17}} ⋅ {\sqrt{17} \over \sqrt{17}} = {7 \sqrt{17} \over 17} = \frac{7}{17} \sqrt{17} \text{.}\)

1p

1p

d

\(\sqrt{\frac{7}{13}}\)

BreukInWortel (4)
008e - Wortels vereenvoudigen - basis - 0ms

d

\(\sqrt{\frac{7}{13}} = {\sqrt{7} \over \sqrt{13}} ⋅ {\sqrt{13} \over \sqrt{13}} = {\sqrt{91} \over 13} = \frac{1}{13} \sqrt{91} \text{.}\)

1p

opgave 2

Herleid.

1p

a

\({28 \sqrt{168} \over 7 \sqrt{7}}\)

Delen (4)
00dc - Wortels vereenvoudigen - basis - 7ms

a

\({28 \sqrt{168} \over 7 \sqrt{7}} = {28 \over 7} ⋅ {\sqrt{168} \over \sqrt{7}} = 4 \sqrt{24} = 4 ⋅ \sqrt{4} ⋅ \sqrt{6} = 4 ⋅ 2 ⋅ \sqrt{6} = 8 \sqrt{6}\)

1p

1p

b

\(5 \sqrt{2} ⋅ 3 \sqrt{10}\)

Vermenigvuldigen (5)
00dd - Wortels vereenvoudigen - basis - 2ms - data pool: #22 (2ms)

b

\(5 \sqrt{2} ⋅ 3 \sqrt{10} = 15 \sqrt{20} = 15 ⋅ \sqrt{4} ⋅ \sqrt{5} = 15 ⋅ 2 ⋅ \sqrt{5} = 30 \sqrt{5}\)

1p

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