4 – Breuken met letters 4.1 – Optellen en aftrekken 1 De onderstaande breuken zijn al gelijknamig dus je kunt ze direct optellen of aftrekken: a 3 5 3+5 8 + = = p p p p b a b a+b + = 2x 2x 2x c 4 a 4+a + = a +1 a +1 a +1 d 1 2 x +1 − x 2 x +1 = 1− x x2 + 1 e 3 + q 2 + q 3 + q − (2 + q) 3 + q − 2 − q 1 − = = = p p p p p f 3+ a 3 a 1 − = = 2a 2a 2a 2 g 2a − 1 3 2a + 2 2( a + 1) + = = =2 a +1 a +1 a +1 a +1 h 2 x + 1 2 − ( x + 1) 1 − x − = = x+3 x+3 x+3 x+3 i 3q + 4 2 − q 3q + 4 − (2 − q) 4q + 2 − = = 1 − 2q 1 − 2 q 1 − 2q 1 − 2q 2 Eerst maak je de breuken gelijknamig, daarna kun je ze optellen of aftrekken: a 2 5 2 ⋅ b 5 ⋅ a 5a + 2b + = + = a b a ⋅b b ⋅a ab b 3 2 3 ⋅ q 2 ⋅ p 3q − 2 p − = − = p q p⋅q q⋅ p pq c d 2+ p 2 ⋅ q p p + 2q = + = q q q q a+ a ⋅ (a + b) a a a 2 + ab + a = + = a+b a+b a+b a+b e 2 2 ⋅ ( x − 1) 1 1⋅ x 2x − 2 x 2 x − 2 + x 3x − 2 + = + = 2 + 2 = = 2 x x − 1 x ⋅ ( x − 1) ( x − 1) ⋅ x x − x x − x x2 − x x −x f a −1 a − 1 2 ⋅ ( a + 1) a − 1 2a + 2 3a + 1 +2= + = + = a +1 a +1 a +1 a +1 a +1 a +1 g 1 ⋅ ( 2 x + 3) x 1 x⋅2 2x 2x + 3 −3 − = − = − = 2 x + 3 2 ( 2 x + 3) ⋅ 2 2 ⋅ ( 2 x + 3) 4 x + 6 4 x + 6 4 x + 6 h y x2 − 3xy y ⋅ 5 3xy ⋅ x 2 5 y 3x 3 y 5 y − 3x 3 y = 2 − = 2− = 5 x ⋅5 5 ⋅ x2 5x 5 x2 5x 2 © Noordhoff Uitgevers Uitwerkingen 1 4 – Breuken met letters i a ⋅ (a − b) b ⋅ (a + b) a b a 2 − ab ab + b2 a 2 + b2 + = + = 2 + = a + b a − b ( a + b ) ⋅ ( a − b ) ( a − b ) ⋅ ( a + b ) a − b2 a 2 − b2 a 2 − b2 j 1 + a 1 − a (1 + a ) ⋅ ( a + b ) (1 − a ) ⋅ ( a − b ) a + b + a 2 + ab a − b − a 2 + ab 2a 2 + 2b − = − = − = 2 a − b a + b (a − b) ⋅ (a + b) (a + b) ⋅ (a − b) a2 − b2 a 2 − b2 a − b2 k 1 1 ab b a a + b + ab + = + + = a b ab ab ab ab 1+ l a + 1 a − 1 ( a + 1)(a + 1) ( a − 1)( a − 1) a 2 + 2a + 1 − ( a 2 − 2a + 1) 4a − = − = = 2 a − 1 a + 1 ( a − 1)( a + 1) (a + 1)( a − 1) ( a + 1)(a − 1) a −1 m x x ( x − 2) x2 + 4 2 2( x + 2) + = + = 2 x + 2 x − 2 ( x + 2)( x − 2) ( x − 2)( x + 2) x − 4 n x+6 x x ( x + 3) x 2 + 3x − 18 − ( x 2 + 3 x) ( x + 6)( x − 3) 18 − = − = =− 2 x + 3 x − 3 ( x + 3)( x − 3) ( x − 3)( x + 3) ( x + 3)( x − 3) x −9 o x + 1 1 ( x + 1) x 2 y x3 + x2 + y + 2= + = y x yx 2 x2 y x2 y 3 Eerst maak je de breuken gelijknamig (dit kan op een zuinige manier, want er zijn gemeenschappelijke factoren). Daarna kun je ze optellen of aftrekken: a a c a⋅d c⋅c ad + c 2 + = + = bc bd bc ⋅ d bd ⋅ c bcd b − 2 3 2⋅ z 3 ⋅ x 3x − 2 z + =− + = xy yz xy ⋅ z yz ⋅ x xyz c 4a b2 2a b 2 2a ⋅ 3a b 2 ⋅ b 6a 2 + b3 + = + = + = 2bc 3ac bc 3ac bc ⋅ 3a 3ac ⋅ b 3abc d 1 x − 1 1⋅ x x − 1 2 x − 1 + 2 = + 2 = 2 x x⋅x x x x e 2x 2 3a b f g + 3y 4ab 2 = 2 x ⋅ 4b 2 3a b ⋅ 4b + 3 y ⋅ 3a 2 4ab ⋅ 3a = 8 xb + 9 ya 12a 2b 2 1 1 1⋅ y 1⋅ x x+ y 1 + = + = = x ( x + y ) y ( x + y ) x ( x + y ) ⋅ y y ( x + y ) ⋅ x xy ( x + y ) xy 1 1 + = 1 ⋅ 6b2 c 4 4 a 5 b 2 c 6a 3 b 4 c 5 4 a 5 b 2 c ⋅ 6b 2 c 4 6b 2 c 4 + 4a 2 2a 2 + 3b 2 c 4 = = 24a5b4 c5 12a5b4 c 5 + 1 ⋅ 4a 2 6a3b 4 c5 ⋅ 4a 2 = 6b 2 c 4 24a5b4 c 5 + 4a 2 24a5b4 c5 h r q p r⋅r q ⋅q p⋅ p r2 q2 p2 p2 + q2 + r 2 + + = + + = + + = pq pr qr pq ⋅ r pr ⋅ q qr ⋅ p pqr pqr pqr pqr i 2 xy 2 z − 3 x3 y + © Noordhoff Uitgevers 1 x2 z3 = 2 ⋅ x2 z3 xy 2 z ⋅ x 2 z 3 − 3 ⋅ yz 3 x3 y ⋅ yz 3 + 1 ⋅ xy 2 x 2 z 3 ⋅ xy 2 = 2 x 2 z 3 − 3 yz 3 + xy 2 x3 y 2 z 3 Uitwerkingen 2 4 – Breuken met letters j 1 1 x2 y 1 ⋅ x 1⋅ y 2 x2 y − x + y + 2 = 2⋅ 2 − + 2 = xy x x y xy ⋅ x x ⋅ y x2 y 2− k a b2 c l + b c a2 b3 bc 2 a 2 + b3 + bc 2 + = 2 + 2 + 2 = ac ab ab c ab c ab c ab 2 c x 1 x (a − 1) a +1 ax − x − a − 1 − = − = a( a + 1) a( a − 1) a(a + 1)( a − 1) a(a − 1)(a + 1) a (a + 1)( a − 1) m y − y2 + 3y − 1 1 y (2 − y ) 1− y y2 − 3y + 1 − = − = 2 =− 2 1 − y 2 − y (1 − y )(2 − y ) (2 − y )(1 − y ) y − 3 y + 2 y − 3y + 2 n 1 2 x −4 o − 1 2 x + 2x 3 a2 − a − 2 + = a −1 a2 + a 1 1 x x−2 2 − = − = = 2 ( x + 2)( x − 2) x( x + 2) x( x + 2)( x − 2) x( x + 2)( x − 2) x( x − 4) = 3 a −1 3a ( a − 1)( a − 2) 3a + a 2 − 3a + 2 + = + = = ( a + 1)(a − 2) a( a + 1) a( a + 1)(a − 2) a( a + 1)( a − 2) a( a + 1)( a − 2) a2 + 2 a( a 2 − a − 2) © Noordhoff Uitgevers Uitwerkingen 3
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