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Valor numérico de expresiones compuestas
Ejercicio 12
Para este grupo de
ejercicios se debe reemplazar cada letra por su respectivo valor
numérico y realizar las operaciones aritméticas correspondientes

Hallar, el valor numérico de las expresiones siguientes para: a=3, b=4, c=



1



3


, d=



1



2


, m=6, n=



1



4




  1. a

    2

    –2ab+

    b

    2

    =

    3

    2

    –2(
    3
    )
    (
    4
    )
    +

    4

    2

    =9–24+16=1



  2. c

    2

    +2cd+

    d

    2

    =

    (


    1

    3


    )

    2

    +
    2
    (


    1

    3


    )
    (


    1


    2



    )
    +

    (


    1

    2


    )

    2

    =

    1

    9

    +

    1

    3

    +

    1

    4

    =


    4+12+9


    36

    =

    25

    36



  3. a

    c

    +

    b

    d

    =

    3



    1

    3



    +

    4



    1

    2



    =9+8=17



  4. c

    d

    –

    m

    n

    +2=



    1

    3





    1

    2



    –

    6



    1

    4



    +2=

    2

    3

    –24+2=

    2

    3

    –22=


    2–66


    3

    =–

    64

    3

     ↔ –21

    1

    3






  5. a

    2



    3

    –



    b

    2



    2

    +



    m

    2



    6

    =



    3


    3





    3


    –



    4

    2



    2


    +





    6


    2





    6


    =3–



    2






    2


    +6=3–8+6=1



  6. 3

    5

    c–

    1

    2

    b+2d=


    3


    5

    (


    1


    3



    )
    –

    1


    2


    +
    2
    (


    1


    2



    )
    =

    1

    5

    –2+1=

    1

    5

    –1=


    1–5


    5

    =–

    4

    5




  7. ab


    n

    +


    ac


    d

    –


    bd


    m

    =


    3(
    4
    )



    1

    4



    +



    3
    (


    1


    3



    )



    1

    2



    –


    (


    1


    2



    )

    6

    =



    12



    1

    4



    +

    1



    1

    2



    –


    2




    =48+2–

    1

    3

    =50–

    1

    3

    =


    150–1


    3

    =

    149

    3

     ⟺ 49

    2

    3



  8. b
    +
    n
    +

    6m

    =
    4
    +


    1

    4


    +

    6(
    6
    )

    =


    2


    2





    +


    1



    2


    2







    +


    6


    2





    =2+

    1

    2

    +6=8+

    1

    2

    =


    16+1


    2

    =

    17

    2

     ↔ 8

    1

    2

  9. c

    3a

    –d

    16

    b

    2



    +n

    8d

    =

    1

    3



    3(
    3
    )

    –

    1

    2





    4


    2


    (4

    )


    2






    +

    1

    4




    2
    .

    2


    2


    (


    1


    2



    )



    =

    3

    3

    –




    2


    +

    1



    (

    2

    )
    =1–8+

    1

    2

    =

    1

    2

    –7=


    1–14


    2

    =–

    13

    2

     ↔ –6

    1

    2





  10. m

    a





    d

    b



    =



    6

    3





    (


    1

    2


    )

    4

    =

    216



    1

    16



    =216(
    16
    )
    =3456




  11. 3

    c

    2




    4

    +


    4

    n

    2




    m

    =



    3


    1



    3


    2







    4

    +



    4


    1



    4


    2







    6

    =

    1

    12

    +

    1

    24

    =


    2+1


    24

    =


    3




    =

    1

    8





  12. 4

    d

    2




    2

    +


    16

    n

    2




    2

    –1=



    4
    (


    1




    2

    2





    )

    2

    +



    16
    (


    1




    4

    2





    )

    2

    –1=–1=0




  13. a+b


    c

    –


    b+m


    d

    =


    3+4




    1

    3



    –


    4+6




    1

    2



    =

    7



    1

    3



    –

    10



    1

    2



    =21–20=1




  14. b–a


    n

    +


    m–b


    d

    +5a=


    4–3




    1

    4



    +


    6–4




    1

    2



    +5(
    3
    )
    =

    1



    1

    4



    +

    2



    1

    2



    +15=4+4+15=23




  15. 12c–a



    2b


    –


    16n–a


    m

    +

    1

    d

    =


    (


    1


    3



    )
    –3


    2(
    4
    )

    –




    (


    1


    4



    )
    –3

    6

    +

    1



    1

    2



    =

    1

    8

    –

    1

    6

    +2=


    3–4+48


    24

    =

    47

    24

     ↔ 1

    23

    24



  16. 4b

    +



    3a



    3

    –



    6m



    6

    =

    4(
    4
    )

    +



    3(
    3
    )

    3

    –



    6(
    6
    )

    6

    =


    4


    2





    +




    3


    2







    3

    –




    6


    2







    6

    =4+


    3



    3


    –


    6



    6


    =4+1–1=4





  17. b
    +

    2d




    2

    –




    3c

    +

    8d




    4

    =



    4
    +


    2
    (


    1


    2



    )

    2

    –





    3
    (


    1


    3



    )

    +

    (


    1


    2



    )

    4

    =


    2+1


    2

    –


    1+2


    4

    =

    3

    2

    –



    3

    4

    =


    6–3


    4

    =

    3

    4









  18. 2



    a

    2



    b

    2






    3

    +


    3

    2+

    d

    2






    4

    –a
    n



    =





    2

    (3

    )


    2


    (4

    )


    2









    3

    +


    3

    2+

    (


    1

    2


    )

    2

    4

    –3


    1

    4





    =





    2
    (
    3
    )

    (
    4
    )


    3


    +


    3

    2+

    1

    4






    4

    –3(


    1

    2


    )




    =


    8+


    3



    8+1


    4





    4

    –

    3

    2








    =





    16–3


    2

    +


    3


    9

    4





    4








    =




    13

    2

    +


    3(


    3

    2


    )

    4




    =




    13

    2

    +

    9

    8








    =





    52+9


    8

    =

    61

    8

     ↔ 7

    5

    8




Categories: Preliminares