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PRELIMINARES

Valor numérico de expresiones compuestas

Ejercicio 13

Para este conjunto de
problemas se debe reemplazar cada letra por su respectivo valor numérico
y realizar las operaciones correspondientes

Hallar, el valor numérico de las expresiones siguientes para: a=1, b=2, c=3, d=4, m=

1

2

, n=

2

3

,p=

1

4

, x=0

(
a+b

)
c–d=(
1+2

)
3–4=3(
3
)
–4=9–4=5
(
a+b

)
(
b–a

)
=(
1+2

)
(
2–1

)
=3(
1
)
=3
$\begin{matrix} \end{matrix}$

(
b–m

)
(
c–n

)
+4

a

2

=

(
2–

1

2

)
(
3–

2

3

)
+4(1

)

2

=

(

4–1

2

)
(

9–2

3

)
+4

=

3

2

(

7

3

)
+4

=

7

2

+4=

7+8

2

=

15

2

↔ 7

1

2
$\begin{matrix} \end{matrix}$

(
2m+3n

)
(
4p+

b

2

)
=

(

2

1

2

+
3

2

3

)
(

4

1

4

+

2

2

)

=

(
1+2

)
(
1+4

)

=

3(
5
)

=

15
$\begin{matrix} \end{matrix}$

(
4m+8p

)
(

a

2

+

b

2

)
(
6n–d

)
=

(

1

2

+

1

4

)
(

1

2

+

2

2

)
(

2

3

–4

)

=

=

0
$\begin{matrix} \end{matrix}$

(
c–b

)
(
d–c

)
(
b–a

)
(
m–p

)
=

(
3–2

)
(
4–3

)
(
2–1

)
(

1

2

–

1

4

)

=

(
1
)
(
1
)
(
1
)
(

2–1

4

)

=

1

4
$\begin{matrix} \end{matrix}$

b

2

(
c+d

)
–

a

2

(
m+n

)
+2x
=

2

2

(
3+4

)
–

1

2

(

1

2

+

2

3

)
+2(
0
)

=

4(
7
)
–(

3+4

6

)

=

28–

7

6

=

168–7

6

=

161

6

↔ 26

5

6
$\begin{matrix} \end{matrix}$

2mx+6(

b

2

+

c

2

)
–4

d

2

=

+6(

2

2

+

3

2

)
–4(4

)

2

=

6(
13
)
–64

=

78–64

=

14
$\begin{matrix} \end{matrix}$

(

8m

9n

+

16p

b

)
a
=

(

1

2

2

3

+

1

4

2

)
(
1
)

=

(

+

2

)

=

2

3

+2=

2+6

3

=

8

3

↔ 2

2

3
$\begin{matrix} \end{matrix}$

x+m(

a

b

+

d

c

–

c

a

)
=

0+

1

2

(

1

2

+

4

3

–

3

1

)

=

1

2

(
1+64–3

)

=

2

=

31
$\begin{matrix} \end{matrix}$

4(
m+p

)
a

+

a

2

+

b

2

c

2

=

4(

1

2

+

1

4

)
1

÷

1

2

+

2

2

3

2

=

4
(

2+1

4

)
÷

5

9

=

3 ÷

5

9

↔

3

5

9

=

27

5

↔ 5

2

5
$\begin{matrix} \end{matrix}$

(
2m+3n+4p

)
(
8p+6n–4m

)
(
9n+20p

)
=

(

2

1

2

+
3

2

3

+
4

1

4

)
(

1

4

+

2

3

–

1

2

)
(

2

3

+

1

4

)

=

(
1+2+1

)
(

2
+4–
2

)
(
6+5

)

=

(
4
)
(
4
)
(
11
)

=

176
$\begin{matrix} \end{matrix}$

c

2

(
m+n

)
–

d

2

(
m+p

)
+

b

2

(
n+p

)
=

3

2

(

1

2

+

2

3

)
–

4

2

(

1

2

+

1

4

)
+

2

2

(

2

3

+

1

4

)

=

(

3+4

)
–(

2+1

4

)
+
4
(

8+3

)

=

3(

7

2

)
–4(
3
)
+

11

3

=

21

2

–12+

11

3

=

63–72+22

6

=

13

6

↔ 2

1

6
$\begin{matrix} \end{matrix}$

(

c

2

+

d

2

a

÷

2

d

)
m
=

(

c

2

+

d

2

a

2

d

)
m

=

(

3

2

+

4

2

1

2

4

)

1

2

=

(

25

2

2

)

1

2

=

5

2

↔ 2

1

2
$\begin{matrix} \end{matrix}$

(
4p+2b

)
(
18n–24p

)
+2(
8m+2

)
(
40p+a

)
=

(

4

1

4

+2(
2
)
)
(

2

3

–

1

4

)
+2(

1

2

+2

)
(

1

4

+1

)

=

(
1+4

)
(
12–6

)
+2(
4+2

)
(
10+1

)

=

5(
6
)
+2(
6
)
(
11
)

=

30+132

=

162
$\begin{matrix} \end{matrix}$

a+

d

b

d–b

×

5+

2

m

2

p

2

=

1+

2

4–2

×

5+

2

(

1

2

)
2

(

1

4

)
2

=

3

2

×

5+

2

1

4

1

16

=

3

2

×

5+8

1

16

=

3

2

×

13

1

16

=

3

2

× 13(

=

312
$\begin{matrix} \end{matrix}$

(
a+b

)

c

2

+8b

–m

n

2

=

(
1+2

)

3

2

+8(
2
)
–

1

2

(

2

3

)

2

=

3

9+16

–

1

2

(

2

3

)

=

3
25
–

1

3

=

3(
5
)
–

1

3

=

15–

1

3

=

45–1

3

=

44

3

↔ 14

2

3
$\begin{matrix} \end{matrix}$

(

a+c

2

+

6n

b

)
÷ (
c+d

)
p

=

(

1+3

2

+

2

3

2

)
÷ (
3+4

)

1

4

=

(

4

2

+

4

2

)
÷ 7(

1

2

)

=

(

2

2

+

2

2

)
÷

7

2

=

2

7

2

=

4

7
$\begin{matrix} \end{matrix}$

3(
c–b

)

32m

–2(
d–a

)

16p

–

2

n

=

3(
3–2

)

1

2

–2(
4–1

)

1

4

–

2

2

3

=

3(
1
)
(
4
)
–2(
3
)
(
2
)
–3

=

12
–
12
–3

=

–3
$\begin{matrix} \end{matrix}$

6abc

2

8b

+

3mn

2(
b–a

)
–

cdnp

abc

=

6(
1
)
(
2
)
(
3
)

2

8(
2
)
+

3
(

1

2

)
(

2

3

)

2(
2–1

)
–

3
(
4

)
(

2

3

)
(

1

4

)

(
1
)
(
2

)
(
3
)

=

36

2
16

+

1

2

–

1

3

=

2
(
4
)
+(

3–2

6

)

=

3

4

+

1

6

=

9+2

12

=

11

12
$\begin{matrix} \end{matrix}$

a

2

+

b

2

b

2

–

a

2

+3(
a+b

)
(
2a+3b

)
=

1

2

+

2

2

2

2

–

1

2

+3(
1+2

)
(
2 × 1+3 × 2

)

=

1+4

4–1

+3(
3
)
(
2+6

)

=

5

3

+72=

5+216

3

=

221

3

↔ 73

2

3
$\begin{matrix} \end{matrix}$

b

2

+(

1

a

+

1

b

)
(

1

b

+

1

c

)
+

(

1

n

+

1

m

)
2

=

2

2

+(

1

1

+

1

2

)
(

1

2

+

1

3

)
+

(

1

2

3

+

1

1

2

)
2

=

4+(

2+1

2

)
(

3+2

6

)
+

(

3

2

+2

)
2

=

4+(

3

2

)
(

5

)
+

(

3+4

2

)
2

=

4+

5

4

+

(

7

2

)
2

=

(

16+5

4

)
+

49

4

=

21

4

+

49

4

=

21+49

4

=

35

2

↔ 17

1

2
$\begin{matrix} \end{matrix}$

(
2m+3n

)
(
4p+2c

)
–4

m

2

n

2

=

(

2

1

2

+
3

2

3

)
(

4

1

4

+2 × 3

)
–4

(

1

2

)
2

(

2

3

)
2

=

(
1+2

)
(
1+6

)
–
4
(

1

4

)
(

4

9

)

=

3(
7
)
–

4

9

=

21–

4

9

=

189–4

9

=

185

9

↔ 20

5

9
$\begin{matrix} \end{matrix}$

b

2

–

c

3

2ab–m

–

n

b–m

=

2

2

–

3

3

2(
1
)
(
2
)
–

1

2

–

2

3

2–

1

2

=

4–1

4–

1

2

–

2

3

4–1

2

=

3

8–1

2

–

2

3

3

2

=

3

7

2

–

4

9

=

6

7

–

4

9

=

54–28

63

=

26

63