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PRELIMINARES

Reducción de un polinomio que contenga términos semejantes de diversas clases

Ejercicio 10

Para resolver este tipo de
problemas se tiene que agrupar todos los terminos de la misma clase, se
reducen los términos de cada clase y obtenemos el polinomio resultante

7a–9b+6a–4b=(
–9b–4b

)
+(
7a+6a

)
=13a–13b
a+b–c–b–c+2c–a=++=0
5x–11y–9+20x–1–y=(
5x+20x

)
+(
–11y–y

)
+(
–9–1

)
=25x–12y–10
–6m+8n+5–m–n–6m–11=(
–6m–m

)
+(
8n–n

)
+(
5–11

)
=–7m+7n–6
–a+b+2b–2c+3a+2c–3b=(
–a+3a

)
++=2a
–81x+19y–30z+6y+80x+x–25y=+–30z=–30z
15

a

2

–6ab–8

a

2

+20–5ab–31+

a

2

–ab=(
15

a

2

–8

a

2

+

a

2

)
+(
–6ab–ab–5ab

)
+(
20–31

)
=8

a

2

–12ab–11
–3a+4b–6a+81b–114b+31a–a–b=(
–3a–6a+31a–a

)
+(
4b+81b–114b–b

)
=21a–30
–71

a

3

b–84

a

4

b

2

+50

a

3

b+84

a

4

b

2

–45

a

3

b+18

a

3

b=+(
–71

a

3

b+50

a

3

b–45

a

3

b+18

a

3

b

)
=–48

a

3

b
–a+b–c+8+2a+2b–19–2c–3a–3–3b+3c=+++(
8–19–3

)
=–14
m

2

+71mn–14

m

2

–65mn+

m

3

–

m

2

–115

m

2

+6

m

3

=(

m

3

+6

m

3

)
+(

m

2

–14

m

2

–

m

2

–115

m

2

)
+(
71mn–65mn

)
=7

m

3

–129

m

2

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

x

4

y–

x

3

y

2

+

x

2

y–8

x

4

y–

x

2

y–10+

x

3

y

2

–7

x

3

y

2

–9+21

x

4

y–

y

3

+50

=

(

x

4

y–8

x

4

y+21

x

4

y

)
+(
–

x

3

y

2

+

x

3

y

2

–7

x

3

y

2

)
+(

x

2

y

–

x

2

y

)
–

y

3

+(
–10–9+50

)
=

14

x

4

y–7

x

3

y

2

–

y

3

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

5

a

x+1

–3

b

x+2

–8

c

x+3

–5

a

x+1

–50+4

b

x+2

–65–

b

x+2

+90+

c

x+3

+7

c

x+3

=

+++(
–50–65+90

)
=

–25
$\begin{matrix} \end{matrix}$

a

m+2

–

x

m+3

–5+8–3

a

m+2

+5

x

m+3

–6+

a

m+2

–5

x

m+3

=

(

a

m+2

–3

a

m+2

+

a

m+2

)
+(
5

x

m+3

–

x

m+3

–5

x

m+3

)
+(
–5+8–6

)
=

–

a

m+2

–

x

m+3

–3
$\begin{matrix} \end{matrix}$

0.3a+0.4b+0.5c–0.6a–0.7b–0.9c+3a–3b–3c

=

(
0.3a–0.6a+3a

)
+(
0.4b–0.7b–3b

)
+(
0.5c–0.9c–3c

)
=

2.7a–3.3b–3.4c
$\begin{matrix} \end{matrix}$

1

2

a+

1

3

b+2a–3b–

3

4

a–

1

6

b+

3

4

–

1

2

=

(

1

2

a+2a–

3

4

a

)
+(

1

3

b–3b–

1

6

b

)
+(

3

4

–

1

2

)

=

2a+8a–3a

4

+

2b–18b–b

6

+

3–2

4

=

7a

4

–

17b

6

–

1

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

3

5

m

2

–2mn+

1

10

m

2

–

1

3

mn+2mn–2

m

2

=

(

3

5

m

2

+

1

10

m

2

–2

m

2

)
+(

–2mn

–

1

3

mn+

2mn

)
=

6

m

2

+

m

2

–20

m

2

10

–

1

3

mn

=

–

m

2

–

1

3

mn

=

–

7

m

2

5

–

1

3

mn
$\begin{matrix} \end{matrix}$

–

3

4

a

2

+

1

10

ab–

5

6

b

2

+2

1

3

a

2

–

3

4

ab+

1

6

b

2

–

1

3

b

2

–2ab

=

(
–

3

4

a

2

+

7

3

a

2

)
+(

1

10

ab–

3

4

ab–2ab

)
+(
–

5

6

b

2

+

1

6

b

2

–

1

3

b

2

)
=

–9

a

2

+28

a

2

12

+

2ab–15ab–40ab

20

+

–5

b

2

+

b

2

–2

b

2

6

=

19

a

2

12

–

53ab

20

–

b

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

0.4

x

2

y+31+

3

8

x

y

2

–0.6

y

3

–

2

5

x

2

y–0.2x

y

2

+

1

4

y

3

–6

=

(
–

3

5

y

3

+

1

4

y

3

)
++(

3

8

x

y

2

–

1

5

x

y

2

)
+(
31–6

)
=

–12

y

3

+5

y

3

20

+

15x

y

2

–8x

y

2

40

+25

=

–

7

y

3

20

+

7x

y

2

40

+25
$\begin{matrix} \end{matrix}$

3

25

a

m–1

–

7

50

b

m–2

+

3

5

a

m–1

–

1

25

b

m–2

–0.2

a

m–1

+

1

5

b

m–2

=

(

3

25

a

m–1

+

3

5

a

m–1

–

1

5

a

m–1

)
+(
–

7

50

b

m–2

–

1

25

b

m–2

+

1

5

b

m–2

)
=

3

a

m–1

+15

a

m–1

–5

a

m–1

25

+

–7

b

m–2

–2

b

m–2

+10

b

m–2

50

=

13

a

m–1

25

+

b

m–2

50