226 226 4 1 1 105 2 1 226 216 2 2 1 5 221 2 x a x x x x x x 4 2 1 6 2 1 4 12 4

226 226 4 1 1 105 2 1 226 216 2 2 1 5 221 2 x a x x x

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· 226 226 4 1 1 105 2 1 226 216 2 2 1 5 221 2 x = a) x x x x x x + + = = + + + = − 4 2 1 6 2 1 4 12 4 36 41 12 2 ( ) ( x x x + + = 4 226 1 105 0 2 2 ) . 2 1 3 3 5 0 x x + + = 4 x x x + + = + 12 4 8 x x 2 2 2 4 = 3 x x + + = 4 1 6 2 014 z z z = − ± = − ± = − = 3 3 4 1 4 2 1 3 5 2 1 4 2 1 2 · ( ) · · ( ) b) 4 3 0 3 4 0 4 2 2 4 2 x x x x x + = + + = x x x = − − ± = ± = = − ( ) ( ) · · ( ) · 1 1 4 1 6 2 1 1 5 2 3 2 2 1 2 a) x x x x x x + + = + + = 1 1 2 7 1 6 0 2 4 3 0 4 2 2 x x x + = x x x x + + = + + 1 1 7 1 2 013 2 SOLUCIONARIO
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54 La solución es x = 4. La solución es x = 3. Estas ecuaciones aparecen factorizadas. Encuentra su solución. a) 3( x 1)( x + 2)( x 4) = 0 d) 2 x 2 ( x 3) 2 (3 x + 4) = 0 b) x ( x 2)( x + 3)( x 12) = 0 e) 5 x ( x 1) 2 (2 x + 7) 3 = 0 c) (2 x 1)(4 x + 3)( x 2) = 0 a) x 1 = 1 x 2 = − 2 x 3 = 4 d) x 1 = 0 x 2 = 3 x 3 = b) x 1 = 0 x 2 = 2 x 3 = − 3 x 4 = 12 e) x 1 = 0 x 2 = 1 x 3 = c) x 1 = x 2 = x 3 = 2 Factoriza las ecuaciones y resuélvelas. a) x 4 2 x 3 13 x 2 + 38 x 24 = 0 b) x 5 6 x 4 + 10 x 3 6 x 2 + 9 x = 0 c) x 4 + 8 x 3 + 17 x 2 + 8 x + 16 = 0 a) ( x 1)( x 2)( x 3)( x + 4) = 0 x 1 = 1 x 2 = 2 x 3 = 3 x 4 = − 4 b) x ( x 3) 2 ( x 2 + 1) = 0 x 1 = 0 x 2 = 3 c) ( x + 4) 2 ( x 2 + 1) = 0 x = − 4 Escribe una ecuación que tenga como soluciones: x = 3, x = 2 y x = − 7. ¿Cuál es el mínimo grado que puede tener? Respuesta abierta. ( x 3)( x 2)( x + 7) = 0 El mínimo grado que puede tener es 3. 017 016 3 4 1 2 7 2 4 3 015 x x = − − ± = ± ( ) ( ) · · · 249 249 4 64 171 2 64 249 135 256 2 1 2 57 64 3 = = x d) 2 1 3 4 3 5 0 2 1 3 4 3 5 6 32 2 2 x x x x x + + = + = ( ) ( ) ( ) 2 2 2 30 4 3 64 249 171 0 = − + = ( ) x x x x x x = − − ± = ± = = ( ) ( ) · · · 9 9 4 2 4 2 2 9 7 4 1 2 4 2 1 2 c) x x x x x x x x x x + + = + + + + + = + + = 12 8 4 2 12 12 8 4 4 48 3 2 2 6 96 64 2 9 4 0 2 2 x x x x + + = Ecuaciones, inecuaciones y sistemas
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55 Resuelve estos sistemas de ecuaciones. Escoge el método que consideres más adecuado. a) Resolvemos el sistema por sustitución: b) Resolvemos el sistema por reducción: Sumamos las ecuaciones: Sustituimos en una de las ecuaciones: Halla las soluciones de estos sistemas. a) b) a) Resolvemos el sistema por sustitución: Resolvemos por reducción: Sustituimos en una de las ecuaciones: 2 x + 3 = 2 x = 1 2 + = = − = − = 2 2 2 5 14 4 12 3 x y x y y y 3 2 1 6 4 15 3 2 6 4 ( ) ( ) ( ) x y x y x y x y + = + + + = + = = − 2 2 2 5 14 x y x y x y x y + = + + = 5 3 1 5 2 1 36 2 3(2 4( ) ) 3(2 6(4 3( 2 x y x y x y x y + = + + + = 1 15 6 4 ) ) ) 019 5 2 1 2 1 2 5 p p + = = · = − = = = − 15 6 3 15 10 11 16 8 1 2 p q p q q q 5 2 1 15 10 11 15 6 3 15 1 3 p q p q p q p + = = = − ⎯→ 0 11 q = 6 3 2 9 6 1 3 3 1 3 2 1 2 · · = − = = = − y y y x 4 6 0 6 9 6 3 2 x y x y x y + = = −
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