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Uitwerkingen! Hoofdstuk 4 Basisboek wiskunde Craats en Bosch

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Volledige uitwerkingen en antwoorden van het hoofdstuk 4 uit Basisboek wiskunde Jan Craats en Rob Bosch

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Hoofdstuk 4

Vul de gegeven waarden in, in de gegeven algebraïsche uitdrukking en bereken het resultaat.

4.1
Substitueer a = 3

A.
2𝑎 2 = 2𝑥(32 ) = 18

B.
−𝑎 2 + 𝑎 = −(32 ) + 3 = −6

C.
4𝑎 3 − 2𝑎 = 4𝑥(33 ) − 2𝑥3 = 4𝑥27 − 2𝑥3 = 102

D.
−3𝑎 3 − 3𝑎 2 = −3𝑥(33 ) − 3𝑥(32 ) = −3𝑥27 − 3𝑥9 = −108

E.
𝑎(2𝑎 − 3) = 3((2𝑥3) − 3) = 3𝑥(6 − 3) = 9


4.2
Substitueer a = -2

A.
3𝑎 2 = 3𝑥(−22 ) = 12

B.
−𝑎 3 + 𝑎 = −(−23 ) − 2 = 8 − 2 = 6

C.
3(𝑎 2 − 2𝑎) = 3((−22 ) − (2𝑥 − 2)) = 3(4 − −4) = 3 𝑥 8 = 24

D.
−2𝑎 2 + 𝑎 = −2𝑥(−22 ) − 2 = (−2 𝑥 4) − 2 = −10

E.
2𝑎(−𝑎 + 3) = (2 𝑥 − 2)𝑥(− − 2 + 3) = −4 𝑥 5 = −20


4.3
Substitueer a = 4

A.
3𝑎 2 − 2𝑎 = (3𝑥42 ) − (2𝑥4) = 40

B.
−𝑎 3 + 2𝑎 2 = −(43 ) + (2 𝑥 42 ) = −64 + 32 = −32

, C.
−2(𝑎 2 − 2𝑎) = −2(42 − (2𝑥4)) = −2(16 − 8) = −16
D.
(2𝑎 − 4)(−𝑎 + 2) = ((2𝑥4) − 4)𝑥(−4 + 2) = 4 𝑥 − 2 = −8

E.
(3𝑎 − 4)2 = ((3𝑥4) − 4)2 = 82 = 64

4.4
Substitueer a = -3

A.
−𝑎 2 + 2𝑎 = −(−32 ) + (2 𝑥 − 3) = −(9) + (−6) = −15

B.
𝑎 3 − 2𝑎 2 = (−33 ) − 2𝑥(−32 ) = −27 − 2𝑥(9) = −27 − 18 = −45

C.
−3(𝑎 2 − 2𝑎) = −3((−32 ) − (2 𝑥 − 3)) = −3(9 − −6) = −3 𝑥 15 = −45

D.
(2𝑎 − 1)(−3𝑎 + 2) = ((2𝑥 − 3) − 1)𝑥 ((−3𝑥 − 3) + 2) = (−6 − 1)𝑥(9 + 2) = −77

E.
(2𝑎 + 1)2 = ((2𝑥 − 3) + 1)2 = (−6 + 1)2 = −52 = 25

4.5
Substitueer a = 3 en b=2

A.
2𝑎 2 𝑏 = 2𝑥32 𝑥2 = 36

B.
3𝑎 2 𝑏 2 − 2𝑎𝑏 = 3𝑥32 𝑥22 − 2𝑥3𝑥2 = 108 − 12 = 96

C.
−3𝑎 2 𝑏3 + 2𝑎𝑏2 = −3𝑥32 𝑥23 + 2𝑥3𝑥22 = −216 + 24 = −192

D.
2𝑎 3 𝑏 − 3𝑎𝑏3 = 2𝑥33 𝑥2 − 3𝑥3𝑥23 = 108 − 72 = 36

E.
−5𝑎𝑏2 − 2𝑎 2 +3𝑏 3 = −5𝑥3𝑥22 − 2𝑥32 +3𝑥23 = −60 − 18 + 24 = −54


4.6
Substitueer a =-2 en b =-3

A.
3a𝑏 − 𝑎 = (3 𝑥 − 2 𝑥 − 3) − −2 = 18 + 2 = 20
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