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Ejercicio 126

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CAPITULO XIV

Operaciones con Fracciones
Ejercicio 126
Simplificar:
  1. x–2 4 + 3x+2 6 = 3( x–2 ) +2( 3x+2 ) 12 = 3x–6+6x+4 12 = 9x–2 12
  2. 2 5 a 2 + 1 3ab = 2( 3b ) +5a 15 a 2 b = 6b+5a 15 a 2 b
  3. a–2b 15a + b–a 20b = 4b( a–2b ) +3a( b–a ) 60ab = 4ab–8 b 2 +3ab–3 a 2 60ab = –8 b 2 +7ab–3 a 2 60ab
  4. a+3b 3ab + a 2 b–4a b 2 5 a 2 b 2 = a+3b 3ab + ab ( a–4b ) 5 a 2 b 2 = 5( a+3b ) +3( a–4b ) 15ab = 5a+15b+3a–12b 15ab = 8a+3b 15ab
  5. a–1 3 + 2a 6 + 3a+4 12 = 1 3 [ a–1+ 2 a 2 + 3a+4 4 ] = 1 3 [ a–1+a+ 3a+4 4 ] = 1 3 [ 2a–1+ 3a+4 4 ] = 1 3 [ 4( 2a–1 ) +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. 1–x 2x + x+2 x 2 + 1 3a x 2 = 1 x [ 1–x 2 + x+2 x + 1 3ax ] = 1 x [ 3ax( 1–x ) +6a( x+2 ) +2 6ax ] = 1 x [ 3ax–3a x 2 +6ax+12a+2 6ax ] = 9ax–3a x 2 +12a+2 6a x 2
  8. 2a–3 3a + 3x+2 10x + x–a 5ax = 10x( 2a–3 ) +3a( 3x+2 ) +6( x–a ) 30ax = 20ax–30x+9ax+ 6a +6x– 6a 30ax = 29ax–24x 30ax = x ( 29a–24 ) 30a x = 29a–24 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. x–y 12 + 2x+y 15 + y–4x 30 = 1 3 [ x–y 4 + 2x+y 5 + y–4x 10 ] = 1 3 [ 5( x–y ) +4( 2x+y ) +2( y–4x ) 20 ] = 1 3 [ 5x–5y+ 8x +4y+2y– 8x 20 ] = 1 3 [ 5x+y 20 ] = 5x+y 60
  11. m–n mn + n–a na + 2a–m am = a( m–n ) +m( n–a ) +n( 2a–m ) 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+10–5 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 + 2a–3m am + 3 a = 1 a [ a+3b b + 2a–3m m +3 ] = 1 a [ m( a+3b ) +b( 2a–3m ) +3bm bm ] = 1 a [ am+3bm+2ab– 3bm + 3bm bm ] = 1 a [ am+2ab+3bm bm ] = am+2ab+3bm abm
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