Getal & Ruimte (12e editie) - havo wiskunde B
'Breuken herleiden'.
| 2 havo/vwo | 1.2 Breuken optellen |
opgave 1Herleid tot één breuk. 1p a \({3 \over 6 x} - {8 \over 6 x}\) Optellen (1) 008u - Breuken herleiden - basis - 0ms - dynamic variables a \({3 \over 6 x} - {8 \over 6 x} = -{5 \over 6 x}\) 1p 1p b \({9 \over a} - {8 \over 4 a}\) Optellen (2) 008v - Breuken herleiden - basis - 0ms - dynamic variables b \({9 \over a} - {8 \over 4 a} = {36 \over 4 a} - {8 \over 4 a} = {28 \over 4 a} = {7 \over a}\) 1p 1p c \({7 \over 5 x} - {6 \over 3 y}\) Optellen (3) 008w - Breuken herleiden - basis - 0ms - dynamic variables c \({7 \over 5 x} - {6 \over 3 y} = {21 y \over 15 x y} - {30 x \over 15 x y} = {21 y - 30 x \over 15 x y} = {7 y - 10 x \over 5 x y}\) 1p 1p d \(7 - {4 \over 3 p}\) Optellen (4) 008x - Breuken herleiden - basis - 0ms - dynamic variables d \(7 - {4 \over 3 p} = {7 \over 1} - {4 \over 3 p} = {21 p \over 3 p} - {4 \over 3 p} = {21 p - 4 \over 3 p}\) 1p opgave 2Herleid tot één breuk. 1p a \(2 a + {8 \over 9 a}\) Optellen (5) 008y - Breuken herleiden - basis - 0ms - dynamic variables a \(2 a + {8 \over 9 a} = {2 a \over 1} ⋅ {9 a \over 9 a} + {8 \over 9 a} = {18 a^{2} \over 9 a} + {8 \over 9 a} = {18 a^{2} + 8 \over 9 a}\) 1p 1p b \({7 a \over b} - {3 \over 9 b}\) Optellen (6) 008z - Breuken herleiden - basis - 0ms - dynamic variables b \({7 a \over b} - {3 \over 9 b} = {63 a \over 9 b} - {3 \over 9 b} = {63 a - 3 \over 9 b} = {21 a - 1 \over 3 b}\) 1p 1p c \({2 b \over 5 a} - {6 a \over 8 b}\) Optellen (7) 0090 - Breuken herleiden - basis - 0ms - dynamic variables c \({2 b \over 5 a} - {6 a \over 8 b} = {16 b^{2} \over 40 a b} - {30 a^{2} \over 40 a b} = {-30 a^{2} + 16 b^{2} \over 40 a b} = {-15 a^{2} + 8 b^{2} \over 20 a b}\) 1p opgave 3Herleid. 1p a \({4 p \over p}\) Vereenvoudigen (1) 00h5 - Breuken herleiden - basis - 0ms - dynamic variables a \({4 p \over p} = {4 \over 1} = 4\) 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 \({24 x \over 27 x}\) Vereenvoudigen (3) 00h7 - Breuken herleiden - basis - 0ms - dynamic variables c \({24 x \over 27 x} = \frac{8}{9}\) 1p 1p d \({-18 x \over 2 x}\) Vereenvoudigen (4) 00h8 - Breuken herleiden - basis - 0ms - dynamic variables d \({-18 x \over 2 x} = -9\) 1p opgave 4Herleid. 1p a \({15 x y \over -18 x z}\) Vereenvoudigen (5) 00h9 - Breuken herleiden - basis - 0ms - dynamic variables a \({15 x y \over -18 x z} = -{5 y \over 6 z}\) 1p 1p b \({-21 b \over -24 a b}\) Vereenvoudigen (6) 00ha - Breuken herleiden - basis - 0ms - dynamic variables b \({-21 b \over -24 a b} = {7 \over 8 a}\) 1p 1p c \({-35 p q r \over 5 q r}\) Vereenvoudigen (7) 00hb - Breuken herleiden - basis - 0ms - dynamic variables c \({-35 p q r \over 5 q r} = -7 p\) 1p 1p d \({7 a b \over b} + {2 a c \over c}\) Vereenvoudigen (8) 00hc - Breuken herleiden - basis - 0ms - dynamic variables d \({7 a b \over b} + {2 a c \over c} = 7 a + 2 a = 9 a\) 1p |
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| 2 havo/vwo | 1.3 Breuken vermenigvuldigen en delen |
opgave 1Herleid tot één breuk. 1p a \({6 \over p} ⋅ {5 \over q}\) Vermenigvuldiging (1) 0091 - Breuken herleiden - basis - 0ms - dynamic variables a \({6 \over p} ⋅ {5 \over q} = {30 \over p q}\) 1p 1p b \({x \over 4} ⋅ -{7 \over y}\) Vermenigvuldiging (2) 0092 - Breuken herleiden - basis - 0ms - dynamic variables b \({x \over 4} ⋅ -{7 \over y} = -{7 x \over 4 y}\) 1p 1p c \({8 \over 5} ⋅ a\) Vermenigvuldiging (3) 0093 - Breuken herleiden - basis - 0ms - dynamic variables c \({8 \over 5} ⋅ a = {8 a \over 5}\) 1p 1p d \({4 \over a} : {3 \over b}\) Deling (1) 0095 - Breuken herleiden - basis - 0ms - dynamic variables d \({4 \over a} : {3 \over b} = {4 \over a} ⋅ {b \over 3} = {4 b \over 3 a}\) 1p opgave 2Herleid tot één breuk. 1p \({6 \over 5} : x\) Deling (2) 0096 - Breuken herleiden - basis - 0ms - dynamic variables ○ \({6 \over 5} : x = {6 \over 5} : {x \over 1} = {6 \over 5} ⋅ {1 \over x} = {6 \over 5 x}\) 1p |
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| 3 havo | 5.2 Breuken met letters herleiden |
opgave 1Herleid tot één breuk. 1p \({a \over 4} + {a + 3 \over 9}\) Optellen (8) 0098 - Breuken herleiden - basis - 0ms - dynamic variables ○ \({a \over 4} + {a + 3 \over 9} = {9 a \over 36} + {4 (a + 3) \over 36} = {9 a + 4 (a + 3) \over 36} = {13 a + 12 \over 36}\) 1p |
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| havo wiskunde B | 11.5 Gebroken functies |
opgave 1Herleid tot één breuk. 1p \({-3 a - 8 \over -2 a + 1} - 5\) Optellen (9) 00eh - Breuken herleiden - basis - 0ms - dynamic variables ○ \({-3 a - 8 \over -2 a + 1} - 5 = {-3 a - 8 \over -2 a + 1} + {-5 (-2 a + 1) \over -2 a + 1} = {-3 a - 8 - 5 (-2 a + 1) \over -2 a + 1} = {-3 a - 8 + 10 a - 5 \over -2 a + 1} = {7 a - 13 \over -2 a + 1}\) 1p opgave 2Deel uit. 1p a \({6 x^{2} + 9 x - 30 \over 3 x}\) Uitdelen (1) 00ei - Breuken herleiden - basis - 0ms - dynamic variables a \({6 x^{2} + 9 x - 30 \over 3 x} = {6 x^{2} \over 3 x} + {9 x \over 3 x} - {30 \over 3 x} = 2 x + 3 - {10 \over x}\) 1p 1p b \({9 a^{2} + 7 a - 4 \over 8 a^{2}}\) Uitdelen (2) 00ej - Breuken herleiden - basis - 0ms - dynamic variables b \({9 a^{2} + 7 a - 4 \over 8 a^{2}} = {9 a^{2} \over 8 a^{2}} + {7 a \over 8 a^{2}} - {4 \over 8 a^{2}} = 1\frac{1}{8} + {7 \over 8 a} - {1 \over 2 a^{2}}\) 1p |