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CAPITULO XIV

Miscelania sobre Fracciones
Ejercicio 140
Simplificar:
  1. 12 x 2 +31x+20 18 x 2 +21x4 = 12 x 2 +16x+15x+20 18 x 2 3x+24x4 = 4x(3x+4 ) +5(3x+4 ) 3x(6x1 ) +4(6x1 ) = (4x+5 )(3x+4 ) (3x+4 )(6x1 ) = 4x+5 6x1
  2. ( 1 a + 2 a 2 + 1 a 3 ) ÷ (a+2 2a+1 a ) = 1 a (1+ 2 a + 1 a 2 ) ÷ [ a 2 +2a(2a+1 ) a ] = 1 a ( a 2 +2a+1 a 2 ) ÷ ( a 2 + 2a 2a 1 a ) = 1 a [ (a+1 ) 2 a 2 ] ÷ ( a 2 1 a ) = 1 a [ (a+1 ) 2 a 2 ] × [ a (a+1 )(a1 ) ] = a+1 a 2 (a1 )
  3. x 3 +3 x 2 +9x x 5 27 x 2 = x ( x 2 +3x+9 ) x 2 ( x 3 27 ) = x 2 +3x+9 x(x3 )( x 2 +3x+9 ) = 1 x(x3 )
  4. (x+y ) 2 y x (xy ) 2 xy = 1 y [ (x+y ) 2 x (xy ) 2 x ] = 1 y [ (x+y ) 2 (xy ) 2 ] = 1 y {[(x+y ) +(xy ) ] [(x+y ) (xy ) ] } = 1 y (x+ y +x y ) ( x +y x +y ) = 1 y (2x ) (2 y ) =4x
  5. a 4 2 b 3 + a 2 b(b2 ) a 4 a 2 b2 b 2 = a 4 2 b 3 + a 2 b 2 2 a 2 b a 4 + a 2 b2 a 2 b2 b 2 = a 4 + a 2 b 2 2 a 2 b2 b 3 a 2 ( a 2 +b ) 2b( a 2 +b ) = a 2 ( a 2 + b 2 ) 2b( a 2 + b 2 ) ( a 2 2b ) ( a 2 +b ) = ( a 2 2b )( a 2 + b 2 ) ( a 2 2b )( a 2 +b ) = a 2 + b 2 a 2 +b
  6. Multiplicar a+ 1+5a a 2 5 por a a+5 a+1
    (a+ 1+5a a 2 5 ) (a a+5 a+1 ) =[ a( a 2 5 ) +1+5a a 2 5 ] [ a(a+1 ) (a+5 ) a+1 ] =( a 3 5a +1+ 5a a 2 5 ) ( a 2 + a a 5 a+1 ) =( a 3 +1 a 2 5 ) ( a 2 5 a+1 ) = (a+1 )( a 2 a+1 ) a+1 = a 2 a+1
  7. Dividir x 2 +5x4 x 3 29 x5 entre x+34+ 170 x 2 x5
    ( x 2 +5x4 x 3 29 x5 ) ÷ (x+34+ 170 x 2 x5 ) =[ (x5 ) ( x 2 +5x4 ) ( x 3 29 ) x5 ] ÷ [ (x5 ) (x+34 ) +170 x 2 x5 ] =( x 3 + 5 x 2 4x 5 x 2 25x+20 x 3 +29 x5 ) ÷ ( x 2 +29x 170 + 170 x 2 x5 ) =( 4929x x5 ) ÷ ( 29x x5 ) =( 4929x x5 ) ( x5 29x ) = 4929x 29x
    Descomponer las expresiones siguientes en la suma o resta de tres fracciones simples irreducibles
  8. 4 x 2 5xy+ y 2 3x = 4 x 2 3 x 5 x y 3 x + y 2 3x = 4x 3 5y 3 + y 2 3x
  9. mnx mnx = m m nx n m n x x mn x = 1 nx 1 mx 1 mn
  10. Probar que x 3 x y 2 xy = x 2 +xy
    x 3 x y 2 xy = x 2 +xy x( x 2 y 2 ) xy =x(x+y ) x(x+y )(xy ) xy =x(x+y ) x(x+y ) =x(x+y )
  11. Probar que x 2 2x+1 9x3 x 2 x3 = x 3 1 x1
    x 2 2x+1 9x3 x 2 x3 = x 3 1 x1 (x3 ) ( x 2 2x+1 ) (9x3 x 2 ) x3 = (x1 )( x 2 +x+1 ) x1 (x3 ) ( x 2 2x+1 ) +3x(x3 ) x3 = x 2 +x+1 (x3 )[( x 2 2x+1 ) +3x ] x3 = x 2 +x+1 x 2 2x+1+3x = x 2 +x+1 x 2 +x+1 = x 2 +x+1
  12. Probar que a 4 5 a 2 +4 a 3 + a 2 4a4 =a3+ 2+4a 2a+1
    a 4 5 a 2 +4 a 3 + a 2 4a4 =a3+ 2+4a 2a+1 a 4 a 2 4 a 2 +4 a 2 (a+1 ) 4(a+1 ) = (2a+1 ) (a3 ) +2+4a 2a+1 a 2 ( a 2 1 ) 4( a 2 1 ) ( a 2 4 ) (a+1 ) = 2 a 2 +a6a3+2+4a 2a+1 ( a 2 4 )( a 2 1 ) ( a 2 4 )(a+1 ) = 2 a 2 a1 2a+1 (a+1 )(a1 ) a+1 = 2 a 2 2a+a1 2a+1 a1 = 2a(a1 ) +(a1 ) 2a+1 a1 = (2a+1 )(a1 ) 2a+1 a1 =a1
    Simplificar:
  13. 1 ab + 1 a+b + 2a a 2 ab+ b 2 = (a+b ) ( a 2 ab+ b 2 ) +(ab ) ( a 2 ab+ b 2 ) +2a(a+b ) (ab ) (ab ) (a+b ) ( a 2 ab+ b 2 ) = a 3 a 2 b+a b 2 + a 2 b a b 2 + b 3 + a 3 a 2 b+a b 2 a 2 b + a b 2 b 3 +2a( a 2 b 2 ) (ab ) ( a 3 + b 3 ) = 2 a 3 2 a 2 b+2a b 2 +2a( a 2 b 2 ) (ab ) ( a 3 + b 3 ) = 2a[ a 2 ab+ b 2 + a 2 b 2 ] (ab ) ( a 3 + b 3 ) = 2a[2 a 2 ab ] (ab ) ( a 3 + b 3 ) = 2 a 2 (2ab ) (ab ) ( a 3 + b 3 )
  14. ( a 2 1 a 2 a 4 1 a 4 ) × (1a+ 1+ a 3 a 2 ) =[ a 2 1 a 2 a 4 (1 a 2 ) (1+ a 2 ) ] × [ a 2 (1a ) +1+ a 3 a 2 ] = a 2 1 a 2 [1 a 2 1+ a 2 ] × [ a 2 a 3 +1+ a 3 a 2 ] = 1 1 a 2 ( 1+ a 2 a 2 1+ a 2 ) × ( a 2 +1 ) = 1 1 a 2
  15. ( x 2 9 x 2 x12 ÷ x3 x 2 +3x ) × a 2 x 2 16 a 2 2 x 2 +7x+3 × ( 2 a 2 x + 1 a 2 x 2 ) =[ (x3 ) (x+3 ) (x4 )(x+3 ) × x(x+3 ) x3 ] × a 2 ( x 2 16 ) 2 x 2 +x+6x+3 × 1 a 2 x (2+ 1 x ) = 1 x4 × x (x+3 ) × a 2 (x4 )(x+4 ) x(2x+1 ) +3(2x+1 ) × 1 a 2 x ( 2x+1 x ) = (x+3 ) × x+4 (x+3 ) (2x+1 ) × ( 2x+1 x ) = x+4 x
  16. Mathematical Equation
  17. 1681 x 2 72 x 2 5x12 = (49x ) (4+9x ) 72 x 2 32x+27x12 = (49x ) (4+9x ) 8x(9x+4 ) 3(9x+4 ) = (49x )(4+9x ) (8x3 )(9x+4 ) = 49x 8x3
  18. ( 1 x 2 x+2 + 3 x+3 ) ÷ ( x x+2 + x x+3 + 6 x 2 +5x+6 ) =[ (x+2 ) (x+3 ) 2x(x+3 ) +3x(x+2 ) x(x+2 ) (x+3 ) ] ÷ ( x x+2 + x x+3 + 6 (x+2 ) (x+3 ) ) =[ x 2 +5x+62 x 2 6x +3 x 2 + 6x x(x+2 ) (x+3 ) ] ÷ [ x(x+3 ) +x(x+2 ) +6 (x+2 ) (x+3 ) ] =[ 2 x 2 +5x+6 x(x+2 ) (x+3 ) ] ÷ [ x 2 +3x+ x 2 +2x+6 (x+2 ) (x+3 ) ] = 2 x 2 +5x+6 x (x+2 ) (x+3 ) × (x+2 ) (x+3 ) 2 x 2 +5x+6 = 1 x
  19. b a 1 b 2 a 2 + 1+ b ab 2 a3b ab = b a a 2 b 2 a 2 + a b b ab 2(ab ) (a3b ) ab = ab a 2 b 2 + a 2a2ba+3b = ab (a+b ) (ab ) + a a+b = a a+b [ b ab +1 ] = a a+b [ b +a b ab ] = a 2 a 2 b 2
  20. 1 3 ( x 2 36 x ÷ x x 2 4 ) × 1 x 36 x × 1 x 4 x = 1 3 [ (x+6 ) (x6 ) x × (x2 ) (x+2 ) x ] × 1 x 2 36 x × 1 x 2 4 x = 1 3 (x+6 ) (x6 ) x × (x2 ) (x+2 ) x × x x 2 36 × x x 2 4 = 1 3
  21. 3a (a2b ) 2 + 5 a5b + 1 a2b 3 a 2 14ab+10 b 2 a 2 4ab+4 b 2 = 3a(a5b ) +5 (a2b ) 2 +(a5b ) (a2b ) (a5b ) (a2b ) 2 3 a 2 14ab+10 b 2 (a2b ) 2 = 3 a 2 15ab+5( a 2 4ab+4 b 2 ) + a 2 7ab+10 b 2 (a5b ) (3 a 2 14ab+10 b 2 ) = 3 a 2 22ab+5 a 2 20ab+20 b 2 + a 2 +10 b 2 (a5b ) (3 a 2 14ab+10 b 2 ) = 9 a 2 42ab+30 b 2 (a5b ) (3 a 2 14ab+10 b 2 ) = 3 (3 a 2 14ab+10 b 2 ) (a5b )(3 a 2 14ab+10 b 2 ) = 3 a5b
  22. x+1 x1 x1 x+1 x1 x+1 + x+1 x1 × x 2 +1 2 a 2 2b ÷ 2x a 2 b = (x+1 ) 2 (x1 ) 2 (x1 ) (x+1 ) (x1 ) 2 + (x+1 ) 2 (x1 ) (x+1 ) × x 2 +1 2 ( a 2 b ) × a 2 b 2x = (x+1 ) 2 (x1 ) 2 (x1 ) 2 + (x+1 ) 2 × x 2 +1 2 × 1 2x = [(x+1 ) +(x1 ) ] [(x+1 ) (x1 ) ] x 2 2x +1+ x 2 + 2x +1 × x 2 +1 4x = (x+ 1 +x 1 ) ( x +1 x +1 ) 2 x 2 +2 × x 2 +1 4x = 2(2x ) 2 ( x 2 +1 ) × x 2 +1 4x = 1 2
  23. 1 3x9 1 6x+12 1 2 (x3 ) 2 + 1 x6+ 9 x = 1 3(x3 ) 1 6(x+2 ) 1 2 (x3 ) 2 + 1 x(x6 ) +9 x = 1 3(x3 ) 1 6(x+2 ) 1 2 (x3 ) 2 + 1 x 2 6x+9 x = 1 3(x3 ) 1 6(x+2 ) 1 2 (x3 ) 2 + x x 2 6x+9 = 1 3(x3 ) 1 6(x+2 ) 1 2 (x3 ) 2 + x (x3 ) 2 = 2(x+2 ) (x3 ) (x3 ) 2 3(x+2 ) +6x(x+2 ) 6(x+2 ) (x3 ) 2 = 2( x 2 x6 ) ( x 2 6x+9 ) 3x6+6 x 2 +12x 6(x+2 ) (x3 ) 2 = 2 x 2 2x12 x 2 +6x96+6 x 2 +9x 6(x+2 ) (x3 ) 2 = 7 x 2 +13x27 6(x+2 ) (x3 ) 2
  24. ab+ a 2 + b 2 a+b a+b a 2 2 b 2 ab × b+ b 2 a ab × 1 1+ 2ab b = (a+b ) (ab ) + a 2 + b 2 a+b (ab ) (a+b ) ( a 2 2 b 2 ) ab × ab+ b 2 a ab × 1 b +2a b b = a 2 b 2 + a 2 + b 2 a+b a 2 b 2 a 2 +2 b 2 ab × b(a+b ) a ab × b 2a = 2 a 2 a+b b 2 ab × b(a+b ) a(ab ) × b 2a = 2 a 2 (ab ) b 2 (a+b ) × b (a+b ) (ab ) × b 2 a =1