Getal & Ruimte (13e editie) - havo wiskunde B

'Wortels vereenvoudigen'.

2 havo/vwo 5.3 Wortels herleiden

Wortels vereenvoudigen (4)

opgave 1

Herleid.

1p

a

\(\sqrt{500}\)

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

a

\(\sqrt{500} = \sqrt{100} ⋅ \sqrt{5} = 10 \sqrt{5} \text{.}\)

1p

1p

b

\(-5 \sqrt{50}\)

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

b

\(-5 \sqrt{50} = -5 ⋅ \sqrt{25} ⋅ \sqrt{2} = -5 ⋅ 5 ⋅ \sqrt{2} = -25 \sqrt{2} \text{.}\)

1p

1p

c

\(\sqrt{\frac{1}{100}}\)

BreukInWortel (1)
008b - Wortels vereenvoudigen - basis - 47ms

c

\(\sqrt{\frac{1}{100}} = {\sqrt{1} \over \sqrt{100}} = \frac{1}{10} \text{.}\)

1p

1p

d

\({36 \sqrt{120} \over 4 \sqrt{6}}\)

Delen (4)
00dc - Wortels vereenvoudigen - basis - 9ms

d

\({36 \sqrt{120} \over 4 \sqrt{6}} = {36 \over 4} ⋅ {\sqrt{120} \over \sqrt{6}} = 9 \sqrt{20} = 9 ⋅ \sqrt{4} ⋅ \sqrt{5} = 9 ⋅ 2 ⋅ \sqrt{5} = 18 \sqrt{5}\)

1p

3 havo 5.4 Wortels herleiden

Wortels vereenvoudigen (1)

opgave 1

Herleid.

1p

\(\sqrt{5\frac{13}{16}}\)

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

\(\sqrt{5\frac{13}{16}} = \sqrt{\frac{93}{16}} = {\sqrt{93} \over \sqrt{16}} = {\sqrt{93} \over 4} = \frac{1}{4} \sqrt{93} \text{.}\)

1p

havo wiskunde B 7.2 Afstanden bij punten en lijnen

Wortels vereenvoudigen (6)

opgave 1

Herleid.

2p

a

\(\sqrt{20} + \sqrt{125}\)

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

a

\(\sqrt{20} + \sqrt{125} = \sqrt{4} ⋅ \sqrt{5} + \sqrt{25} ⋅ \sqrt{5} = 2 \sqrt{5} + 5 \sqrt{5} \text{.}\)

1p

\(2 \sqrt{5} + 5 \sqrt{5} = 7 \sqrt{5} \text{.}\)

1p

2p

b

\(7 \sqrt{80} - 4 \sqrt{20}\)

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

b

\(7 \sqrt{80} - 4 \sqrt{20} = 7 ⋅ \sqrt{16} ⋅ \sqrt{5} - 4 ⋅ \sqrt{4} ⋅ \sqrt{5} \text{.}\)

1p

\(7 ⋅ 4 ⋅ \sqrt{5} - 4 ⋅ 2 ⋅ \sqrt{5} = 28 \sqrt{5} - 8 \sqrt{5} = 20 \sqrt{5} \text{.}\)

1p

1p

c

\({8 \over 7 \sqrt{7}}\)

WortelInNoemer
0089 - Wortels vereenvoudigen - basis - 1ms

c

\({8 \over 7 \sqrt{7}} = {8 \over 7 \sqrt{7}} ⋅ {\sqrt{7} \over \sqrt{7}} = {8 \sqrt{7} \over 7 ⋅ 7} = \frac{8}{49} \sqrt{7} \text{.}\)

1p

1p

d

\(\sqrt{3\frac{4}{15}}\)

BreukInWortel (3)
008d - Wortels vereenvoudigen - basis - 1ms

d

\(\sqrt{3\frac{4}{15}} = \sqrt{\frac{49}{15}} = {\sqrt{49} \over \sqrt{15}} = {7 \over \sqrt{15}} ⋅ {\sqrt{15} \over \sqrt{15}} = {7 \sqrt{15} \over 15} = \frac{7}{15} \sqrt{15} \text{.}\)

1p

opgave 2

Herleid.

1p

a

\(\sqrt{\frac{3}{13}}\)

BreukInWortel (4)
008e - Wortels vereenvoudigen - basis - 1ms

a

\(\sqrt{\frac{3}{13}} = {\sqrt{3} \over \sqrt{13}} ⋅ {\sqrt{13} \over \sqrt{13}} = {\sqrt{39} \over 13} = \frac{1}{13} \sqrt{39} \text{.}\)

1p

1p

b

\(2 \sqrt{2} ⋅ 5 \sqrt{14}\)

Vermenigvuldigen (5)
00dd - Wortels vereenvoudigen - basis - 3ms - data pool: #22 (2ms)

b

\(2 \sqrt{2} ⋅ 5 \sqrt{14} = 10 \sqrt{28} = 10 ⋅ \sqrt{4} ⋅ \sqrt{7} = 10 ⋅ 2 ⋅ \sqrt{7} = 20 \sqrt{7}\)

1p

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