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МатАрхива
GeoMentor
Игор Богданоски
Архива
Вештини
Теореми
Автор
Наставници
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Colorblind RG
Colorblind BY
Архива
6 одделение
ai-1752d933
Архива: | 6 одделение | 3/10 тежина
Текст на задачата
Задача:
None
Решение
Прикажи решение
Решение:
Да ги разгледаме комбинациите на броевите и операциите: -
(
−
2
+
3
)
⋅
(
−
4
)
=
1
⋅
(
−
4
)
=
−
4
(-2 + 3) \cdot (-4) = 1 \cdot (-4) = -4
(
−
2
+
3
)
⋅
(
−
4
)
=
1
⋅
(
−
4
)
=
−
4
-
(
−
2
−
3
)
⋅
(
−
4
)
=
−
5
⋅
(
−
4
)
=
20
(-2 - 3) \cdot (-4) = -5 \cdot (-4) = 20
(
−
2
−
3
)
⋅
(
−
4
)
=
−
5
⋅
(
−
4
)
=
20
-
3
⋅
(
−
2
−
(
−
4
)
)
=
3
⋅
2
=
6
3 \cdot (-2 - (-4)) = 3 \cdot 2 = 6
3
⋅
(
−
2
−
(
−
4
))
=
3
⋅
2
=
6
-
(
−
4
⋅
3
)
−
(
−
2
)
=
−
12
+
2
=
−
10
(-4 \cdot 3) - (-2) = -12 + 2 = -10
(
−
4
⋅
3
)
−
(
−
2
)
=
−
12
+
2
=
−
10
-
(
−
4
−
(
−
2
)
)
⋅
3
=
−
2
⋅
3
=
−
6
(-4 - (-2)) \cdot 3 = -2 \cdot 3 = -6
(
−
4
−
(
−
2
))
⋅
3
=
−
2
⋅
3
=
−
6
Најголемата вредност е **20**, добиена како
(
−
2
−
3
)
⋅
(
−
4
)
(-2 - 3) \cdot (-4)
(
−
2
−
3
)
⋅
(
−
4
)
.