a. 9-[(240+220)·(19-19)] = 9
b. 29-20+14-19 =4
c. (-25-17+40)+(35-16-20)= 3
d. 2+5+8-7-3 = 7
e.3·4+(6+5-13)-12/4 = 2
f.(-25+18-45+12)/(26-20+46-32) = 1
g. 35-24-35+23 =1
h.(2·4++10)/(6-3) =6
i.(2·4-4)·(6-4) =8
a. 9-[(240+220)·(19-19)].
9-[460x0] =
9-0 =
9
b. 29-20+14-19=
9 - 5=
4
c. (-25-17+40)+(35-16-20)
-2+(-1) =
-2-1=
-3
d. 2+5+8-7-3 =
5
e.3·4+(6+5-13)-12/4=
3x4+(-2)-12/4=
12-2-12/4=
10-12/4=
28/4 =
7
f.(-25+18-45+12)/(26-20+46-32)
-40/20=
-2
g. 35-24-35+23
-1
h.(2·4++10)/(6-3)
(8+10)/3=
18/3 =
6
i.(2·4-4)·(6-4)=
(8-4)x2=
4x2 =
8
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El valor absoluto de las operaciones indicadas es:
a. 9-[(240+220)·(19-19)] = 9
b. 29-20+14-19 =4
c. (-25-17+40)+(35-16-20)= 3
d. 2+5+8-7-3 = 7
e.3·4+(6+5-13)-12/4 = 2
f.(-25+18-45+12)/(26-20+46-32) = 1
g. 35-24-35+23 =1
h.(2·4++10)/(6-3) =6
i.(2·4-4)·(6-4) =8
Explicación paso a paso:
Al Resolver los Ejercicios tenemos que el valor absoluto del resultado va a ser:
a. 9-[(240+220)·(19-19)].
9-[460x0] =
9-0 =
9
b. 29-20+14-19=
9 - 5=
4
c. (-25-17+40)+(35-16-20)
-2+(-1) =
-2-1=
-3
d. 2+5+8-7-3 =
5
e.3·4+(6+5-13)-12/4=
3x4+(-2)-12/4=
12-2-12/4=
10-12/4=
28/4 =
7
f.(-25+18-45+12)/(26-20+46-32)
-40/20=
-2
g. 35-24-35+23
-1
h.(2·4++10)/(6-3)
(8+10)/3=
18/3 =
6
i.(2·4-4)·(6-4)=
(8-4)x2=
4x2 =
8