1 / 10

Clase 14

Clase 14. Ejercicios sobre ecuaciones con radicales. x – 1. x + 1. √. x – 1 + x – 4. = x – 5. =. √. √. x – 1. x + 2. Para el estudio individual. Determina el conjunto solución de las siguientes ecuaciones:. a). S = { 7 }. b). S = { 5 }. (x – 4 ) √ x – 1.

Download Presentation

Clase 14

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Clase 14 Ejercicios sobre ecuaciones con radicales

  2. x – 1 x + 1 √ x – 1 + x – 4 = x – 5 = √ √ x – 1 x +2 Para el estudio individual Determina el conjunto solución de las siguientes ecuaciones: a) S = {7 } b) S = {5 }

  3. (x – 4) √ x – 1 √ x – 1 ● x + 1 √ x – 1 + x – 4 = √ x – 1 = x – 1 + (x – 4) √ x – 1 2 2 2 = (x – 4)2 (x – 1 ) 4 = 4 = (x2 – 8x + 16)(x – 1) 4 = x3–8x2 +16x –x2 +8x –16 0 = x3–9x2 +24x –20

  4. x3–9x2 +24x –20 = 0 = 0 x – 5 = 0 ó x – 2 = 0 x2 = 2 x1 = 5 1 – 9 24 –20 –20 20 5 5 0 – 4 4 1 (x – 5)(x – 2)2

  5. Ejercicio Resuelve las siguientes ecuaciones: a) √ x + 5 + 1 = x b) x –√9 + x √x2 – 3 = 3

  6. a) √ x + 5 + 1 = x 2 √ x + 5 = x – 1 ( )2 x + 5 = x2 – 2x + 1 x2 – 3x – 4=0 (x – 4)(x + 1) = 0 x – 4= 0ó x + 1= 0 x1= 4 x2= – 1

  7. = √9 +1 = √4 +1 MI:√4 + 5 + 1 MI:√–1 + 5 + 1 Comprobación para x1= 4 = 3+1 = 4 MD:4 comparación:4 = 4 para x2= –1 = 2+1 = 3 MD:–1 comparación:3 ≠ –1

  8. = 9 + x√x2 – 3 = x√x2 – 3 b) x –√9 + x√x2 – 3 = 3 2 x – 3 = √9 + x√x2 – 3 ( )2 x2 – 6x + 9 2 ( )2 x2 – 6x x4–12x3+36x2 = x2(x2 – 3) x4–12x3+36x2 = x4– 3x2 12x3 – 39x2 = 0

  9. 13 x2 = 4 12x3 – 39x2 = 0 3x2 (4x – 13) = 0 x1 = 0 ó 4x – 13 = 0

  10. √2 – x = 1 – √x – 1 15 A(x) + = 8 A(x) Para el estudio individual 1. Determina el conjunto solución de la siguiente ecuación: Resp. S = 1  3 2.Sea A(x) = √x2 + 9 . Determina para que valores de x se cumple: Respuestax = 4 ; 0

More Related