LÍDERES EN REFORZAMIENTO ACADÉMICO

ECA Estudio y Centro de Aprendizajeecacentrodeaprendizaje@gmail.com

martes, 23 de agosto de 2011

CUBO DEL BINOMIO

CUBO DEL BINOMIO

    1. (a + 3)³= a³ + 3(a)²(3) + 3(a)(3)² + (3)³
    = 27 + 9a² + 27a + 27

  1. (p – q)³ = p³ – 3(p)²(q) + 3(p)(q)² – q³
    = p³ – 3p²q + 3pq² - q³

    3. (x + 2)³ = x³ + 3(x)²(2) + 3(x)(2)² + 2³
    = x³ + 6x² + 12x + 8

  1. (a – 3)³ = a³ + 3(a)²(3) + 3(a)(3)²+ (3)³
    = a³ + 9a² + 27a + 27

  1. (t + 4)³ = t³ + 3(t)²(4) + 3(t)(4)² + (4)³ =
    = t³ + 3(t)²(4) + 3(t)(4)² + (4)³
    = t³ + 12t² + 48t + 64
  1. (2 – a)³ = 2³ – 3(2)²(a) + 3(2)(a)² – a³
    = 8 – 12a + 6a² - a³
  1. (2a – b)³ = (2a)³ -3(2a)²(b)+3(2a)(b)²- b³ =8a³ – 3(4a²)b + 6ab² – b³
    = 8a³ – 12a²b + 6ab² – b³

  1. (3a - 5b)³ = (3a)³-3(3a)²(5b)+3(3a)(5b)²-(5b)³
    = 27a³ - 135a²b + 225ab² - 125b³

9. (2x + 3y)³=(2x)³+3(2x)²(3y)+3(2x)(3y)²+(3y)³ = 8x³ + 36 x²y + 54xy² + 27y³

10. (1 – 3y)³ = (1)³ – 3(1)²(3y)+3(1)(3y)²- (3y)³ = 1 – 9y + 27y² - 27y³

11. (2 + 3t)³ = 2³ + 3 (2)²(3t) + 3(2)(3t)² + (3t)³ = 8 + 36t + 54t² +27t
  1. (3a –2x)³=(3a)³–3(3a)²(2x)+3(3a)(2x)²-(2x)³ = 27a³ – 54a²x + 36ax² – 8x³


13. (5a – 1)³= (5 a)³–3(5a)²(1)+3(5a)(1)² - (1)³ =125 a³ – 75a + 15a - 1

14. (3a²-2a)³=(3a)³–3(3a)²(2a)+3(3a)(2a)²-(2a)² = (x)² - (5x) ² = x ² - 25x ²

  1. (t² + t³)³ =(t²)³+ 3(t²)²(t³) + 3(t²)(t³)² + (t³)³
    = t + 3tt³ + 3t²t + t
    = t⁶ + 3t⁷ + 3t⁸ + t⁹

  1. ( 1 + x)³= (1)³ + 3(1)²(x) + 3(1)(x)² + (x
    = 1 + 3x + 3x + x¹²

  1. (2t–3a²)³=(2t)³–3(2t)²(3a²)+3(2t)(3a²)²– (3a²)³ =
    = 8t³ – 36t²a² + 54ta - 27a
18. (u² +5v)³=(u²)³+3(u²)²(5v)+3(u²)(5v)²+ (5v)³
= u + 15u⁴5v + 75u²v² - 125v³

  1. ( ½ - a)³ = (½ - 3(1/2)²(a) +3(1/2)(a)² – a³
= 1/8 – 3a/4 + 3a²/2 – a³

20. (1/2x + 2y)³=(1/2x)³+3(1/2x)²(2y)+ 3(1/2x)(2y)² + (2y)³
    = x³/8 + 6x²y/4 – 12y²x/2 + 8y³
    = x³/8 + 3x²y/2 – 6y²x + 8y³

    21. (2/3a – 1/3b)³ =
    =(2/3a)³–3(2/3a)²(1/3b)+3(2/3a)(1/3b)²+(1/3b)³
= 8a³/27 – 12a²b/27 + 6ab²/27 – 1/27b³
= 8a³/27 – 4a²b/9 + 2ab²/9 – 1/27b³

22. (5p/2 + 3q/2)³ =
=(5p/2)³+3(5p/2)²(3q/2)+3(5p/2)(3q/2)²+(3q/2)³
= 125p³/8 + 225p²q/8 + 135pq²/8 + 27q³/8

  1. ( 1 m/10 – 1 n /5)³ =
= (1m/10)³–3(1m/10)²(1n/5)+3(1m/10)(1n/5)² – (1n/5)³
= m³/1000 – 3mn/500 + 3mn²/250 – n³/125

  1. (a – a/3)³ = a³–3(a)²(a/3)+3(a)(a/3)² – (a/3)³
= a³ – 3a³/3 + 3a³/9 – a³/27
=27a³ – 27a³ + 9a³ – a³ = 8a³
...............   27..............          . 27

  1. (1t/2 + 2t²)³ =
= (1t/2)³ – 3(1t/2)²(2t²) + 3(1t/2)(2t²)² - (2t²)³ =t³/8 – 6t³/4 + 12t/2 – 8t
=t³/8 – 3t³/2 + 6t⁵ - 8t⁶

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