Ejercicio 126

Comparte esto 👍👍
CAPITULO XIV

Operaciones con Fracciones
Ejercicio 126
Simplificar:
  1. x2 4 + 3x+2 6 = 3(x2 ) +2(3x+2 ) 12 = 3x6+6x+4 12 = 9x2 12
  2. 2 5 a 2 + 1 3ab = 2(3b ) +5a 15 a 2 b = 6b+5a 15 a 2 b
  3. a2b 15a + ba 20b = 4b(a2b ) +3a(ba ) 60ab = 4ab8 b 2 +3ab3 a 2 60ab = 8 b 2 +7ab3 a 2 60ab
  4. a+3b 3ab + a 2 b4a b 2 5 a 2 b 2 = a+3b 3ab + ab (a4b ) 5 a 2 b 2 = 5(a+3b ) +3(a4b ) 15ab = 5a+15b+3a12b 15ab = 8a+3b 15ab
  5. a1 3 + 2a 6 + 3a+4 12 = 1 3 [a1+ 2 a 2 + 3a+4 4 ] = 1 3 [a1+a+ 3a+4 4 ] = 1 3 [2a1+ 3a+4 4 ] = 1 3 [ 4(2a1 ) +3a+4 4 ] = 1 3 [ 8a 4 +3a+ 4 4 ] = 1 3 ( 11a 4 ) 11a 12
  6. n m 2 + 3 mn + 2 m = 1 m [ n m + 3 n +2 ] = 1 m [ n 2 +3m+2mn mn ] = n 2 +2mn+3m m 2 n
  7. 1x 2x + x+2 x 2 + 1 3a x 2 = 1 x [ 1x 2 + x+2 x + 1 3ax ] = 1 x [ 3ax(1x ) +6a(x+2 ) +2 6ax ] = 1 x [ 3ax3a x 2 +6ax+12a+2 6ax ] = 9ax3a x 2 +12a+2 6a x 2
  8. 2a3 3a + 3x+2 10x + xa 5ax = 10x(2a3 ) +3a(3x+2 ) +6(xa ) 30ax = 20ax30x+9ax+ 6a +6x 6a 30ax = 29ax24x 30ax = x (29a24 ) 30a x = 29a24 30a
  9. 3 5 + x+2 2x + x 2 +2 6 x 2 = 3(6 x 2 ) +15x(x+2 ) +5( x 2 +2 ) 30 x 2 = 18 x 2 +15 x 2 +30x+5 x 2 +10 30 x 2 = 38 x 2 +30x+10 30 x 2 = 2 (19 x 2 +15x+5 ) x 2 = 19 x 2 +15x+5 15 x 2
  10. xy 12 + 2x+y 15 + y4x 30 = 1 3 [ xy 4 + 2x+y 5 + y4x 10 ] = 1 3 [ 5(xy ) +4(2x+y ) +2(y4x ) 20 ] = 1 3 [ 5x5y+ 8x +4y+2y 8x 20 ] = 1 3 [ 5x+y 20 ] = 5x+y 60
  11. mn mn + na na + 2am am = a(mn ) +m(na ) +n(2am ) amn = am an+ mn am +2an mn amn = an a m n = 1 m
  12. x+2 3x + x 2 2 5 x 2 + 2 x 3 9 x 3 = 1 x [ x+2 3 + x 2 2 5x + 2 x 3 9 x 2 ] = 1 x [ 15 x 2 (x+2 ) +9x( x 2 2 ) +5(2 x 3 ) 45 x 2 ] = 1 x [ 15 x 3 +30 x 2 +9 x 3 18x+105 x 3 45 x 2 ] = 1 x [ 19 x 3 +30 x 2 18x+10 45 x 2 ] = 19 x 3 +30 x 2 18x+10 45 x 3
  13. 1 ab + b 2 a 2 a b 3 + ab+ b 2 a 2 b 2 = 1 ab + b 2 a 2 a b 3 + b (a+b ) a 2 b 2 = 1 ab [1+ b 2 a 2 b 2 + a+b a ] = 1 ab [ a b 2 +a( b 2 a 2 ) + b 2 (a+b ) a b 2 ] = 1 ab [ a b 2 +a b 2 a 3 +a b 2 + b 3 a b 2 ] = 1 ab [ b 3 +3a b 2 a 3 a b 2 ] = b 3 +3a b 2 a 3 a 2 b 3
  14. a+3b ab + 2a3m am + 3 a = 1 a [ a+3b b + 2a3m m +3 ] = 1 a [ m(a+3b ) +b(2a3m ) +3bm bm ] = 1 a [ am+3bm+2ab 3bm + 3bm bm ] = 1 a [ am+2ab+3bm bm ] = am+2ab+3bm abm