Question

sendo:A:a+b-c,B:a-b-c e C:a-b+c determine:a)A-B b)C-A c)(A+B)-C d)C-(A+B) e) B+(A-C)

Answer 1

A:a+b-c
B:a-b-c
C:a-b+c

a)A-B =
(a+b-c)-(a-b-c) =
a+b-c-a+b+c =
b+b =
2b

b)C-A =
(a-b+c)-(a+b-c) =
a-b+c-a-b+c =
-b+c-b+c =
-2b+2c

c)(A+B)-C =
[(a+b-c)+(a-b-c)]-(a-b+c) =
[a+b-c+a-b-c]-(a-b+c) =
[a-c+a-c]-(a-b+c) =
2a-2c-a+b-c =
a-3c+b

d)C-(A+B) =
(a-b+c)-(2a-2c) =
a-b+c-2a+2c =
-a-b+3c

e)B+(A-C) =
a-b-c+[(a+b-c)-(a-b+c) =
a-b-c+[a+b-c-a+b-c] =
a-b-c+(2b-2c) =
a-b-c+2b-2c =
a+b-3c

Espero ter ajudado, boa sorte com os estudos!

Answer 2

Resposta:

A:a+b-c

B:a-b-c

C:a-b+c

a)A-B =

(a+b-c)-(a-b-c) =

a+b-c-a+b+c =

b+b =

2b

b)C-A =

(a-b+c)-(a+b-c) =

a-b+c-a-b+c =

-b+c-b+c =

-2b+2c

c)(A+B)-C =

[(a+b-c)+(a-b-c)]-(a-b+c) =

[a+b-c+a-b-c]-(a-b+c) =

[a-c+a-c]-(a-b+c) =

2a-2c-a+b-c =

a-3c+b

d)C-(A+B) =

(a-b+c)-(2a-2c) =

a-b+c-2a+2c =

-a-b+3c

e)B+(A-C) =

a-b-c+[(a+b-c)-(a-b+c) =

a-b-c+[a+b-c-a+b-c] =

a-b-c+(2b-2c) =

a-b-c+2b-2c =

a+b-3c

Explicação passo a passo: