Logica

*



TRABAJO ACADEMICO

2012- I | Asignatura: | |
| Docente: | |
| Ciclo del curso: | III | MODULO I |

Datos del Alumno: |
Apellidos y Nombres: | |
Código | |
UDED: | LIMA |












UNIVERSIDAD ALAS PERUanas











INTRODUCCCION



1. Ortografía y Formato (2 puntos)
2. Resolver. (2 puntos)

(a)| (b) | (c) | (d) |
(e) | f) |

A)

x 3x-5.2x-1≤0

x2-x-23x-53x-5x-1≤0

x2-x-6x+103x2-3x-5x+5≤0
x2-7x+103x2-8x+5≤0 P(x)Q(x)≤0 PxQ(x)≤0 Q(x)≠0
I.
3x2-8x+5≠0 no se toma x≠1
x≠3/5

II.
x2-7x+103x2-8x+5≤0
x-2 3x-5
x-5 x-1x-2x-5(x-1)≤0

C.S ]-3/5,1[ U [2,5]≤0 |



3/5 1 2 5

B)
x3-2x2-3x>0
x2x2-x-3>0
x2x-3(x+1)>0
x>0 |



x+1>0
x-1>0

x<-1 |

2x-3>0
2x>3
x>3/2 |C)
x-1 x+x-3x+1≤2

x-1x+1+(x-3)≤2x(x+1)
x2-1+x2-3x≤2x2+2x
x2-1+x2-3x≤2x2+2x
2x2-3x-1-2x2-2x≤0
-5x≤0
5x≥0

x≤0 |





D)
x3 x-1≥x

x3-x2+x (x-1)≥0

(x3-x2+x)x-1≥0
x4-x3-x3+x2+x2-x≥0
x4-2x3+2x2-x≥0xx3-1-2x2(x-1)≥0
xx-1x2+x+1-2x2(x-1)≥0
x-1xx2+x+1-2x2≥0
x-1(x)x2-x+1-2x≥0
x-1(x)x2-3x-1≥0

x=-b±b2-4ac2a
x=3±32-41(-1)2(1)
x=3±132

Entonces:3+132=3,30
3-132=-0,30

x-1(x)x-3+132x-3-132≥0
x=1
x=0
x ε3,30
x ε-0,30

3-132 0 1 3+132


CS=]-∞;3-132] U [0,1[ U [ 3+132;∞[

E)
(x6-x2+7)x (x4-1)7(4x4-3x-1)5≥0 el signo e s al revés ≤

(X⁶+X²+7 )X ______ ≤ 0
(X²+1)⁷ (X+1)⁷(X-1)⁷(X-1)⁵(X+1/4)⁵


(X⁶+X²+7 )X ______ ≤ 0
(X²+1)⁷ (X+1)⁷(X-1)¹²(X+1/4)⁵

-10 1/4 1



F)








(X-1)⁵ (X-2)⁴ (-X+3) ≥ 0 EL GRAFICO TAMBIEN ME LO CAMBIAS PLEASE….
X(X+1) (X²+1)

-1 1 2 3

CS. ]-∞;-1] U [1,2] U [ 3,+∞[



a) y3x-5≤2x-1

* x2-x≤6x-10
* x2-7x+10≤0
* (x-5)(x-2)≤0
* x=5x=2

2
5


C.S. = <2,5>
a. 2x3-x2-3x>0
* x2x3-x-3>0
* x2x-3)(x+1>0
* x=0 x=32 x=-1


*
x=1 -2x
* x=2 6

+ ∞
2

C.S. = <2, +∞)
* c. x-1x+x-3x+1≤2
* x-1x+1+x(x-3)≤2(x)(x+1)
* x2-1+x2-3x≤2x2+2x
0≤5x+1
x=0
x=1


0
+ ∞

C.S.(0,+ ∞)

d. x3x-1≥x
x3≥x2-x
x3-x2+x≥0
x(x2-x+1
x=0

+ ∞
0

C.S. (0, + ∞)

e. (x6+x2+7)(x4-1)7(4x2-3x-1)5≥x
x6+x2+7x≤0(x4-1)7(4x2-3x-1)5
x6+x2+7x≤0
x≤0
x≤-1

1. 0
- ∞


C.S. = <0, -∞)
f. (x-1)5(x-2)4(3-x)x(x+1)(x2+1)≥0
(x-1)5(x-2)4(3-x)≥0(x)(x+1)(x2+1)
≥0

x=0x
x=1
x=2
x=3
x=4x

1
3


C.S. =...