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CLASE 36

3. x. –1. 4. +2. 2. y. x. 5. 2. 3. x. x. 2. –3. y. x. = 7 x. 0. 5 x. 2,1. y. CLASE 36. P( x ). =. 7. ( x  0). Se conoce que un envase de poliestireno tiene forma cúbica, con ( x+ 2) dm de largo y su capacidad se puede expresar como x 3 + 6 x + 68.

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CLASE 36

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  1. 3 x –1 4 +2 2 y x 5 2 3 x x 2 –3 y x = 7x 0 5x 2,1 y CLASE 36 P(x) = 7 (x 0)

  2. Se conoce que un envase de poliestireno tiene forma cúbica, con (x+2) dm de largo y su capacidad se puede expresar como x3+ 6x + 68. ¿Cuántos envases como este se pueden almacenar en un local de 64 m3 de capacidad?

  3. 2 x + 6x +68 3 3 = (x +2) 3 V = l l ? l = x +2 l 2 3 (a + b) (a + b) =(a + b)

  4. 3 a)(x 4) 2 (x + 8x + 16) 2 3 = x + 12x + 48x + 64 2 3 3 4 x x 3 3x4 3 b) (m – 5) 2 12x 2 15m 2 3 3 3 5 m m 3m5 Efectúa y simplifica. 2 (x + 4) 4 x = (x 4) = (x + 4) 48x 64 2 3x 4 + + + 2 = (m – 5) (m – 5) 5 m 2 (m – 5m + 25) = (m – 5) 2 75m 3 = m –15m + 75m –125 125 2 + – 3m 5 –

  5. V = l3 2 x+ 6x +68 3 3 = (x +2) l ? l l =x +2 3 2 2 3 3 – b 3ab + + (a + b) = a + 3a b – –

  6. 3 V = l = x +2 l = 7 l 2 3 3 x + 6x +68 (x +2) = l 2 3 2 x +3x  2 + 3x 4 + 8 3 = x + 6x +68 2 3 2 3 x + 6x + 12x + 8 = x + 6x +68 12x + 8 = – 8 68 :12 12x = 60 x = 5

  7. 3 V = l 3 V = (7,0 dm) 3 V = 343dm 343 l 3 3 1 m = 1000 dm 3 3 64 m = 64000 dm 64000 :  186,5 Rta: En el local se pueden almacenar 186 envases como este.

  8. Convierte en suma 2 a) (x + 2)(x– 2x + 4) 3 = x + 8 3 3 = x + 2 3 3 = y– 3 2 2 2 – 2x + 2x –3y 2 b) (y– 3)(y + 3y + 9) 2 + 3y 3 = y– 27 3 2 2 = x– 2x + 4x + 2x– 4x + 8 – 4x + 4x 2 3 2 – 9y = y+ 3y + 9y–3y– 9y– 27 + 9y 2 3 2 3 (a+b)( a–ab+b) = a+ b +

  9. Productos notables 2 3 2 3 3 3ab b + + (a + b) = a + 3a b 3 2 2 3 3 3ab b – + (a–b) = a – 3a b 2 3 2 3 (a+b)( a–ab + b) = a+ b 3 2 2 3 (a–b)( a+ ab + b) = a – b 2 (x + a) (x + b)=x + (a + b)x +ab 2 2 (a + b)(a– b) = a – b 2 2 (a + b) = a + 2 ab + b2 – –

  10. Estudio Independiente

  11. Realiza las operaciones indicadas y simplifica. 3 a) (2m –n) 3 b) (3a + 2b) 2 3 c) 2(x – 3) –(x – 4) –3 (x + 7) – 4 2 3 d) (2t + 1) – (2 + t)(4 –2t + t ) 2 e) 5(a– 6)(a + 6a + 36) + (a+3)(a –3)

  12. Halla los valores reales de x para los cuales: 3 2 2 (x + 2) – x (x + 16) – (6x –5)= 21

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