Getal & Ruimte (13e editie) - 3 havo
'Breuken herleiden'.
| 2 havo/vwo | 1.2 Breuken optellen |
opgave 1Herleid tot één breuk. 1p a \({2 \over 3 a} + {4 \over 3 a}\) Optellen (1) 008u - Breuken herleiden - basis - 0ms - dynamic variables a \({2 \over 3 a} + {4 \over 3 a} = {6 \over 3 a} = {2 \over a}\) 1p 1p b \({3 \over p} - {5 \over 6 p}\) Optellen (2) 008v - Breuken herleiden - basis - 0ms - dynamic variables b \({3 \over p} - {5 \over 6 p} = {18 \over 6 p} - {5 \over 6 p} = {13 \over 6 p}\) 1p 1p c \({2 \over 5 x} + {7 \over 6 y}\) Optellen (3) 008w - Breuken herleiden - basis - 0ms - dynamic variables c \({2 \over 5 x} + {7 \over 6 y} = {12 y \over 30 x y} + {35 x \over 30 x y} = {12 y + 35 x \over 30 x y}\) 1p 1p d \(3 - {5 \over 9 a}\) Optellen (4) 008x - Breuken herleiden - basis - 0ms - dynamic variables d \(3 - {5 \over 9 a} = {3 \over 1} - {5 \over 9 a} = {27 a \over 9 a} - {5 \over 9 a} = {27 a - 5 \over 9 a}\) 1p opgave 2Herleid tot één breuk. 1p a \(4 x - {2 \over 5 x}\) Optellen (5) 008y - Breuken herleiden - basis - 0ms - dynamic variables a \(4 x - {2 \over 5 x} = {4 x \over 1} ⋅ {5 x \over 5 x} - {2 \over 5 x} = {20 x^{2} \over 5 x} - {2 \over 5 x} = {20 x^{2} - 2 \over 5 x}\) 1p 1p b \({6 a \over b} - {3 \over 5 b}\) Optellen (6) 008z - Breuken herleiden - basis - 0ms - dynamic variables b \({6 a \over b} - {3 \over 5 b} = {30 a \over 5 b} - {3 \over 5 b} = {30 a - 3 \over 5 b}\) 1p 1p c \({4 b \over 3 a} + {9 a \over 2 b}\) Optellen (7) 0090 - Breuken herleiden - basis - 0ms - dynamic variables c \({4 b \over 3 a} + {9 a \over 2 b} = {8 b^{2} \over 6 a b} + {27 a^{2} \over 6 a b} = {27 a^{2} + 8 b^{2} \over 6 a b}\) 1p opgave 3Herleid. 1p a \({9 p \over p}\) Vereenvoudigen (1) 00h5 - Breuken herleiden - basis - 0ms - dynamic variables a \({9 p \over p} = {9 \over 1} = 9\) 1p 1p b \({x \over 6 x}\) Vereenvoudigen (2) 00h6 - Breuken herleiden - basis - 0ms - dynamic variables b \({x \over 6 x} = {1 \over 6}\) 1p 1p c \({-10 x \over -25 x}\) Vereenvoudigen (3) 00h7 - Breuken herleiden - basis - 0ms - dynamic variables c \({-10 x \over -25 x} = \frac{2}{5}\) 1p 1p d \({-18 p \over -2 p}\) Vereenvoudigen (4) 00h8 - Breuken herleiden - basis - 0ms - dynamic variables d \({-18 p \over -2 p} = 9\) 1p opgave 4Herleid. 1p a \({-10 x y \over 14 x z}\) Vereenvoudigen (5) 00h9 - Breuken herleiden - basis - 0ms - dynamic variables a \({-10 x y \over 14 x z} = -{5 y \over 7 z}\) 1p 1p b \({25 b \over -45 a b}\) Vereenvoudigen (6) 00ha - Breuken herleiden - basis - 0ms - dynamic variables b \({25 b \over -45 a b} = -{5 \over 9 a}\) 1p 1p c \({20 x y z \over 4 y z}\) Vereenvoudigen (7) 00hb - Breuken herleiden - basis - 0ms - dynamic variables c \({20 x y z \over 4 y z} = 5 x\) 1p 1p d \({7 a b \over b} - {4 a c \over c}\) Vereenvoudigen (8) 00hc - Breuken herleiden - basis - 0ms - dynamic variables d \({7 a b \over b} - {4 a c \over c} = 7 a - 4 a = 3 a\) 1p |
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| 2 havo/vwo | 1.3 Breuken vermenigvuldigen en delen |
opgave 1Herleid tot één breuk. 1p a \({2 \over x} ⋅ -{3 \over y}\) Vermenigvuldiging (1) 0091 - Breuken herleiden - basis - 0ms - dynamic variables a \({2 \over x} ⋅ -{3 \over y} = -{6 \over x y}\) 1p 1p b \({a \over 9} ⋅ -{3 \over b}\) Vermenigvuldiging (2) 0092 - Breuken herleiden - basis - 0ms - dynamic variables b \({a \over 9} ⋅ -{3 \over b} = -{3 a \over 9 b} = -{a \over 3 b}\) 1p 1p c \({4 \over 3} ⋅ p\) Vermenigvuldiging (3) 0093 - Breuken herleiden - basis - 0ms - dynamic variables c \({4 \over 3} ⋅ p = {4 p \over 3}\) 1p 1p d \({6 \over x} : {2 \over y}\) Deling (1) 0095 - Breuken herleiden - basis - 0ms - dynamic variables d \({6 \over x} : {2 \over y} = {6 \over x} ⋅ {y \over 2} = {6 y \over 2 x} = {3 y \over x}\) 1p opgave 2Herleid tot één breuk. 1p \(-{7 \over 9} : a\) Deling (2) 0096 - Breuken herleiden - basis - 0ms - dynamic variables ○ \(-{7 \over 9} : a = -{7 \over 9} : {a \over 1} = -{7 \over 9} ⋅ {1 \over a} = -{7 \over 9 a}\) 1p |
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| 3 havo | 5.2 Breuken met letters herleiden |
opgave 1Herleid tot één breuk. 1p \({9 p \over 2} + {p + 5 \over 7}\) Optellen (8) 0098 - Breuken herleiden - basis - 0ms - dynamic variables ○ \({9 p \over 2} + {p + 5 \over 7} = {63 p \over 14} + {2 (p + 5) \over 14} = {63 p + 2 (p + 5) \over 14} = {65 p + 10 \over 14}\) 1p |