Getal & Ruimte (13e editie) - 2 havo/vwo

'Breuken herleiden'.

2 havo/vwo 1.2 Breuken optellen

Breuken herleiden (15)

opgave 1

Herleid tot één breuk.

1p

a

\({6 \over 7 a} + {8 \over 7 a}\)

Optellen (1)
008u - Breuken herleiden - basis - 0ms - dynamic variables

a

\({6 \over 7 a} + {8 \over 7 a} = {14 \over 7 a} = {2 \over a}\)

1p

1p

b

\({3 \over x} + {9 \over 4 x}\)

Optellen (2)
008v - Breuken herleiden - basis - 0ms - dynamic variables

b

\({3 \over x} + {9 \over 4 x} = {12 \over 4 x} + {9 \over 4 x} = {21 \over 4 x}\)

1p

1p

c

\({4 \over 5 x} - {8 \over 7 y}\)

Optellen (3)
008w - Breuken herleiden - basis - 0ms - dynamic variables

c

\({4 \over 5 x} - {8 \over 7 y} = {28 y \over 35 x y} - {40 x \over 35 x y} = {28 y - 40 x \over 35 x y}\)

1p

1p

d

\(5 - {2 \over 7 p}\)

Optellen (4)
008x - Breuken herleiden - basis - 0ms - dynamic variables

d

\(5 - {2 \over 7 p} = {5 \over 1} - {2 \over 7 p} = {35 p \over 7 p} - {2 \over 7 p} = {35 p - 2 \over 7 p}\)

1p

opgave 2

Herleid tot één breuk.

1p

a

\(3 a + {7 \over 4 a}\)

Optellen (5)
008y - Breuken herleiden - basis - 0ms - dynamic variables

a

\(3 a + {7 \over 4 a} = {3 a \over 1} ⋅ {4 a \over 4 a} + {7 \over 4 a} = {12 a^{2} \over 4 a} + {7 \over 4 a} = {12 a^{2} + 7 \over 4 a}\)

1p

1p

b

\({2 a \over b} - {5 \over 9 b}\)

Optellen (6)
008z - Breuken herleiden - basis - 0ms - dynamic variables

b

\({2 a \over b} - {5 \over 9 b} = {18 a \over 9 b} - {5 \over 9 b} = {18 a - 5 \over 9 b}\)

1p

1p

c

\({9 y \over 3 x} - {2 x \over 8 y}\)

Optellen (7)
0090 - Breuken herleiden - basis - 0ms - dynamic variables

c

\({9 y \over 3 x} - {2 x \over 8 y} = {72 y^{2} \over 24 x y} - {6 x^{2} \over 24 x y} = {-6 x^{2} + 72 y^{2} \over 24 x y} = {-x^{2} + 12 y^{2} \over 4 x y}\)

1p

opgave 3

Herleid.

1p

a

\({9 a \over a}\)

Vereenvoudigen (1)
00h5 - Breuken herleiden - basis - 0ms - dynamic variables

a

\({9 a \over a} = {9 \over 1} = 9\)

1p

1p

b

\({x \over 8 x}\)

Vereenvoudigen (2)
00h6 - Breuken herleiden - basis - 0ms - dynamic variables

b

\({x \over 8 x} = {1 \over 8}\)

1p

1p

c

\({10 p \over -16 p}\)

Vereenvoudigen (3)
00h7 - Breuken herleiden - basis - 0ms - dynamic variables

c

\({10 p \over -16 p} = -\frac{5}{8}\)

1p

1p

d

\({32 a \over -4 a}\)

Vereenvoudigen (4)
00h8 - Breuken herleiden - basis - 0ms - dynamic variables

d

\({32 a \over -4 a} = -8\)

1p

opgave 4

Herleid.

1p

a

\({20 p q \over 36 p r}\)

Vereenvoudigen (5)
00h9 - Breuken herleiden - basis - 0ms - dynamic variables

a

\({20 p q \over 36 p r} = {5 q \over 9 r}\)

1p

1p

b

\({8 y \over 36 x y}\)

Vereenvoudigen (6)
00ha - Breuken herleiden - basis - 0ms - dynamic variables

b

\({8 y \over 36 x y} = {2 \over 9 x}\)

1p

1p

c

\({15 a b c \over -3 b c}\)

Vereenvoudigen (7)
00hb - Breuken herleiden - basis - 0ms - dynamic variables

c

\({15 a b c \over -3 b c} = -5 a\)

1p

1p

d

\({5 x y \over y} - {6 x z \over z}\)

Vereenvoudigen (8)
00hc - Breuken herleiden - basis - 0ms - dynamic variables

d

\({5 x y \over y} - {6 x z \over z} = 5 x - 6 x = -x\)

1p

2 havo/vwo 1.3 Breuken vermenigvuldigen en delen

Breuken herleiden (5)

opgave 1

Herleid tot één breuk.

1p

a

\({6 \over a} ⋅ {5 \over b}\)

Vermenigvuldiging (1)
0091 - Breuken herleiden - basis - 0ms - dynamic variables

a

\({6 \over a} ⋅ {5 \over b} = {30 \over a b}\)

1p

1p

b

\({x \over 9} ⋅ -{6 \over y}\)

Vermenigvuldiging (2)
0092 - Breuken herleiden - basis - 0ms - dynamic variables

b

\({x \over 9} ⋅ -{6 \over y} = -{6 x \over 9 y} = -{2 x \over 3 y}\)

1p

1p

c

\({7 \over 2} ⋅ p\)

Vermenigvuldiging (3)
0093 - Breuken herleiden - basis - 0ms - dynamic variables

c

\({7 \over 2} ⋅ p = {7 p \over 2}\)

1p

1p

d

\({2 \over a} : {8 \over b}\)

Deling (1)
0095 - Breuken herleiden - basis - 0ms - dynamic variables

d

\({2 \over a} : {8 \over b} = {2 \over a} ⋅ {b \over 8} = {2 b \over 8 a} = {b \over 4 a}\)

1p

opgave 2

Herleid tot één breuk.

1p

\({1 \over 8} : x\)

Deling (2)
0096 - Breuken herleiden - basis - 0ms - dynamic variables

\({1 \over 8} : x = {1 \over 8} : {x \over 1} = {1 \over 8} ⋅ {1 \over x} = {1 \over 8 x}\)

1p

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