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

\({4 \over 6 x} - {7 \over 6 x}\)

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

a

\({4 \over 6 x} - {7 \over 6 x} = -{3 \over 6 x} = -{1 \over 2 x}\)

1p

1p

b

\({2 \over a} + {9 \over 8 a}\)

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

b

\({2 \over a} + {9 \over 8 a} = {16 \over 8 a} + {9 \over 8 a} = {25 \over 8 a}\)

1p

1p

c

\({3 \over 9 p} + {5 \over 2 q}\)

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

c

\({3 \over 9 p} + {5 \over 2 q} = {6 q \over 18 p q} + {45 p \over 18 p q} = {6 q + 45 p \over 18 p q} = {2 q + 15 p \over 6 p q}\)

1p

1p

d

\(5 + {3 \over 7 a}\)

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

d

\(5 + {3 \over 7 a} = {5 \over 1} + {3 \over 7 a} = {35 a \over 7 a} + {3 \over 7 a} = {35 a + 3 \over 7 a}\)

1p

opgave 2

Herleid tot één breuk.

1p

a

\(6 x + {4 \over 7 x}\)

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

a

\(6 x + {4 \over 7 x} = {6 x \over 1} ⋅ {7 x \over 7 x} + {4 \over 7 x} = {42 x^{2} \over 7 x} + {4 \over 7 x} = {42 x^{2} + 4 \over 7 x}\)

1p

1p

b

\({9 x \over y} + {5 \over 6 y}\)

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

b

\({9 x \over y} + {5 \over 6 y} = {54 x \over 6 y} + {5 \over 6 y} = {54 x + 5 \over 6 y}\)

1p

1p

c

\({8 b \over 5 a} - {7 a \over 2 b}\)

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

c

\({8 b \over 5 a} - {7 a \over 2 b} = {16 b^{2} \over 10 a b} - {35 a^{2} \over 10 a b} = {-35 a^{2} + 16 b^{2} \over 10 a b}\)

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

\({x \over 3 x}\)

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

b

\({x \over 3 x} = {1 \over 3}\)

1p

1p

c

\({15 p \over -21 p}\)

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

c

\({15 p \over -21 p} = -\frac{5}{7}\)

1p

1p

d

\({-24 x \over 3 x}\)

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

d

\({-24 x \over 3 x} = -8\)

1p

opgave 4

Herleid.

1p

a

\({35 p q \over -45 p r}\)

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

a

\({35 p q \over -45 p r} = -{7 q \over 9 r}\)

1p

1p

b

\({10 b \over 15 a b}\)

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

b

\({10 b \over 15 a b} = {2 \over 3 a}\)

1p

1p

c

\({-12 a b c \over 2 b c}\)

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

c

\({-12 a b c \over 2 b c} = -6 a\)

1p

1p

d

\({2 x y \over y} + {3 x z \over z}\)

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

d

\({2 x y \over y} + {3 x z \over z} = 2 x + 3 x = 5 x\)

1p

2 havo/vwo 1.3 Breuken vermenigvuldigen en delen

Breuken herleiden (5)

opgave 1

Herleid tot één breuk.

1p

a

\({2 \over a} ⋅ -{3 \over b}\)

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

a

\({2 \over a} ⋅ -{3 \over b} = -{6 \over a b}\)

1p

1p

b

\({p \over 9} ⋅ -{5 \over q}\)

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

b

\({p \over 9} ⋅ -{5 \over q} = -{5 p \over 9 q}\)

1p

1p

c

\(-{1 \over 7} ⋅ x\)

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

c

\(-{1 \over 7} ⋅ x = -{x \over 7}\)

1p

1p

d

\({5 \over x} : {4 \over y}\)

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

d

\({5 \over x} : {4 \over y} = {5 \over x} ⋅ {y \over 4} = {5 y \over 4 x}\)

1p

opgave 2

Herleid tot één breuk.

1p

\({8 \over 9} : a\)

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

\({8 \over 9} : a = {8 \over 9} : {a \over 1} = {8 \over 9} ⋅ {1 \over a} = {8 \over 9 a}\)

1p

"