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Ejercicio 252

Gestor Baldor
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CAPITULO XXXI

R a d i c a l e s
Resolución de Ecuaciones con radicales que se reducen a primer grado
Ejercicio252
Resolver las ecuaciones:
  1. x + x+5 = 10 x x 2 + x( x+5 ) =10 x+ x( x+5 ) =10 ( x( x+5 ) ) 2 = ( 10–x ) 2 x 2 +5x =100–20x+ x 2 5x+20x =100 25x =100 x = 25 x =4
  2. 4x–11 +2 x = 55 4x–11 ( 4x–11 ) 2 +2 x( 4x–11 ) =55 4x–11+2 x( 4x–11 ) =55 2 x( 4x–11 ) =66–4x 2 x( 4x–11 ) = 2 ( 33–2x ) ( x( 4x–11 ) ) 2 = ( 33–2x ) 2 4 x 2 –11x =1089–132x+ 4 x 2 132x–11x =1089 121x =1089 x = 121 x =9
  3. x – x–7 = 4 x x 2 – x( x–7 ) =4 x– x 2 –7x =4 ( x–4 ) 2 = ( x 2 –7x ) 2 x 2 –8x+16 = x 2 –7x –8x+7x =–16 –x =–16 x =16
  4. x –2 x +4 = x +1 x +13 ( x –2 ) ( x +13 ) =( x +1 ) ( x +4 ) x 2 +13 x –2 x –26 = x 2 +4 x + x +4 11 x –5 x =26+4 6 x =30 x = 6 ( x ) 2 = 5 2 x =25
  5. 6 x+8 = x+8 – x 6 = ( x+8 ) 2 – x( x+8 ) 6 =x+8– x 2 +8x x 2 +8x =x+8–6 ( x 2 +8x ) 2 = ( x+2 ) 2 x 2 +8x = x 2 +4x+4 8x–4x =4 4x =4 x = 4 4 x =1
  6. x–3 + 8 x+9 = x+9 ( x–3 ) ( x+9 ) +8 x+9 = x+9 x 2 +6x–27 +8 = ( x+9 ) 2 x 2 +6x–27 +8 =x+9 ( x 2 +6x–27 ) 2 = ( x+1 ) 2 x 2 +6x–27 = x 2 +2x+1 6x–2x =27+1 4x =28 x = 4 x =7
  7. x +4 x –2 = x +11 x –1 ( x +4 ) ( x –1 ) =( x +11 ) ( x –2 ) x 2 – x +4 x –4 = x 2 –2 x +11 x –22 3 x =9 x –22+4 –6 x =–18 x = – – 6 ( x ) 2 = 3 2 x =9
  8. 2 x+6 – 4x–3 = 9 4x–3 2 ( x+6 ) ( 4x–3 ) – ( 4x–3 ) 2 =9 2 4 x 2 –3x+24x–18 –4x+3 =9 2 4 x 2 +21x–18 =9–3+4x 2 ( 4 x 2 +21x–18 ) 2 = 2 ( 2x+3 ) 2 4 x 2 +21x–18 = 4 x 2 +12x+9 21x–12x =18+9 9x =27 x = 9 x =3
  9. x –2 x +2 = 2 x –5 2 x –1 ( x –2 ) ( 2 x –1 ) =( 2 x –5 ) ( x +2 ) 2 x 2 – x – 4 x +2 = 2 x 2 +4 x – 5 x –10 2+10 =4 x 12 =4 x ( 4 ) 2 = ( x ) 2 x =9
  10. x+14 – x–7 = 6 x–7 ( x+14 ) ( x–7 ) – ( x–7 ) 2 =6 x 2 +7x–98 –x+7 =6 x 2 +7x–98 =6–7+x ( x 2 +7x–98 ) 2 = ( x–1 ) 2 x 2 +7x–98 = x 2 –2x+1 7x+2x =98+1 9x =99 x = 9 x =11
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