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Ejercicio 139

Gestor Baldor
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CAPITULO XIV

Operaciones con Fracciones
Ejercicio 139
Hallar el verdadero valor de:
  1. x–2 x+3 parax =2 x–2 x+3 = 2–2 2+3 = 0 5 =0
  2. x–2 x–3 parax =3 x–2 x–3 = 3–2 3–3 = 1 0 = ∞
  3. x 2 – a 2 x 2 + a 2 parax =a x 2 – a 2 x 2 + a 2 = a 2 – a 2 a 2 + a 2 = 0 2 a 2
  4. x 2 + y 2 x 2 – y 2 parax =y x 2 + y 2 x 2 – y 2 = y 2 + y 2 y 2 – y 2 = 2 y 2 0 = ∞
  5. x–1 3 x–2 parax =2 x–1 3 x–2 = ( x–1 ) ( x–2 ) 3 = x 2 –3x+2 3 = 2 2 –3( 2 ) +2 3 = 4 – 6 + 2 3 = 0 3 =0
  6. x 2 –9 x 2 +x–12 parax =3 x 2 –9 x 2 +x–12 = ( x+3 ) ( x–3 ) ( x+4 ) ( x–3 ) = x+3 x+4 = 3+3 3+4 = 6 7
  7. a 2 –a–6 a 2 +2a–15 paraa =3 a 2 –a–6 a 2 +2a–15 = ( a–3 ) ( a+2 ) ( a+5 ) ( a–3 ) = a+2 a+5 = 3+2 3+5 = 5 8
  8. x 2 –7x+10 x 3 –2 x 2 –x+2 parax =2 x 2 –7x+10 x 3 –2 x 2 –x+2 = ( x–5 ) ( x–2 ) x 2 ( x–2 ) –( x–2 ) = ( x–5 ) ( x–2 ) ( x 2 –1 ) ( x–2 ) = x–5 x 2 –1 = 2–5 2 2 –1 =– 3 3 =–1
  9. x 2 –2x+1 x 3 –2 x 2 –x+2 parax =1 x 2 –2x+1 x 3 –2 x 2 –x+2 = ( x–1 ) 2 x 2 ( x–2 ) –( x–2 ) = ( x–1 ) 2 ( x 2 –1 ) ( x–2 ) = ( x–1 ) 2 ( x+1 ) ( x–1 ) ( x–2 ) = x–1 ( x+1 ) ( x–2 ) = 1 – 1 ( 1+1 ) ( 1–2 ) = 0 2( –1 ) =0
  10. a 3 –8 a 2 +11a–26 paraa =2 a 3 –8 a 2 +11a–26 = ( a–2 ) ( a 2 +2a+4 ) ( a+13 ) ( a–2 ) = a 2 +2a+4 a+13 = 2 2 +2( 2 ) +4 2+13 = 4+4+4 2+13 = = 4 5
  11. x 2 –7x+6 x 2 –2x+1 parax =1 x 2 –7x+6 x 2 –2x+1 = ( x–6 ) ( x–1 ) ( x–1 ) 2 = x–6 x–1 = 1–6 1–1 = –5 0 = ∞
  13. x 2 –16 x 3 –4 x 2 –x+4 parax =4 x 2 –16 x 3 –4 x 2 –x+4 = ( x–4 ) ( x+4 ) x 2 ( x–4 ) –( x–4 ) = ( x–4 ) ( x+4 ) ( x 2 –1 ) ( x–4 ) = x+4 ( x–1 ) ( x+1 ) = 4+4 ( 4–1 ) ( 4+1 ) = 8 3( 5 ) = 8 15
  14. 4 x 2 –4x+1 4 x 2 +8x–5 parax = 1 2 4 x 2 –4x+1 4 x 2 +8x–5 = ( 2x–1 ) 2 4 x 2 –2x+10x–5 = ( 2x–1 ) 2 2x( 2x–1 ) +5( 2x–1 ) = ( 2x–1 ) 2 ( 2x+5 ) ( 2x–1 ) = 2x–1 2x+5 = 2 ( 1 2 ) –1 2 ( 1 2 ) +5 = 1–1 1+5 =0
  17. x 3 – a 3 x–a parax =a x 3 – a 3 x–a = ( x–a ) ( x 2 +ax+ a 2 ) ( x–a ) = x 2 +ax+ a 2 = a 2 +a( a ) + a 2 =3 a 2
  18. a 2 –2ab+ b 2 a 2 –ab parab =a a 2 –2ab+ b 2 a 2 –ab = ( a–b ) 2 a ( a–b ) = a–b a = a – a a =0
  19. x 2 – y 2 xy– y 2 y =x x 2 – y 2 xy– y 2 = ( x–y ) ( x+y ) y ( x–y ) = x+y y = x+x x = 2 x x =2
  20. x 3 – a 3 a 2 x– a 3 parax =a x 3 – a 3 a 2 x– a 3 = ( x–a ) ( x 2 +ax+ a 2 ) a 2 ( x–a ) = x 2 +ax+ a 2 a 2 = a 2 +a( a ) + a 2 a 2 = 3 a 2 a 2 =3
  26. 8 x 2 +6x–9 12 x 2 –13x+3 parax = 3 4 8 x 2 +6x–9 12 x 2 –13x+3 = 8 x 2 +12x–6x–9 12 x 2 –4x–9x+3 = 4x( 2x+3 ) –3( 2x+3 ) 4x( 3x–1 ) –3( 3x–1 ) = ( 4x–3 ) ( 2x+3 ) ( 4x–3 ) ( 3x–1 ) = 2x+3 3x–1 = 2 ( 3 ) +3 3( 3 4 ) –1 = 3 2 +3 9 4 –1 = 3+6 2 9–4 = 9 5 2 = 18 5
  27. x 3 +6 x 2 +12x+8 x 4 –8 x 2 +16 parax =–2 x 3 +6 x 2 +12x+8 x 4 –8 x 2 +16 = ( x 3 +8 ) +( 6 x 2 +12x ) ( x 2 –4 ) 2 = ( x+2 ) ( x 2 –2x+4 ) +6x( x+2 ) [ ( x–2 ) ( x+2 ) ] 2 = ( x+2 ) [ ( x 2 –2x+4 ) +6x ] ( x–2 ) 2 ( x+2 ) 2 = x 2 +4x+4 ( x–2 ) 2 ( x+2 ) = ( x+2 ) 2 ( x–2 ) 2 ( x+2 ) = x+2 ( x–2 ) 2 = – 2 + 2 ( –2–2 ) 2 =0
  28. 9 x 3 +3 x 2 +3x+1 27 x 3 +1 parax =– 1 3 9 x 3 +3 x 2 +3x+1 27 x 3 +1 = 3 x 2 ( 3x+1 ) +( 3x+1 ) ( 3x+1 ) ( 9 x 2 –3x+1 ) = ( 3 x 2 +1 ) ( 3x+1 ) ( 3x+1 ) ( 9 x 2 –3x+1 ) = 3 x 2 +1 9 x 2 –3x+1 = 3 ( – 1 3 ) 2 +1 9 ( – 1 3 ) 2 – 3 ( – 1 3 ) +1 = 3 ( 1 3 2 ) +1 9 ( 1 3 2 ) +1+1 = 1 3 +1 3 = 1+3 3 3 = 4 9
  29. 1 x–1 – 3 x 3 –1 parax =1 1 x–1 – 3 x 3 –1 = x 2 +x+1–3 ( x–1 ) ( x 2 +x+1 ) = x 2 +x–2 ( x–1 ) ( x 2 +x+1 ) = ( x+2 ) ( x–1 ) ( x–1 ) ( x 2 +x+1 ) = x+2 x 2 +x+1 = 1+2 1 2 +1+1 = 3 3 =1
  30. ( x 2 +3x–10 ) ( 1+ 1 x–2 ) parax =2 ( x 2 +3x–10 ) ( 1+ 1 x–2 ) =( x+5 ) ( x–2 ) ( x–2+1 x–2 ) =( x+5 ) ( x–1 ) =( 2+5 ) ( 2–1 ) =7( 1 ) =7
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