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Byju's Answer
Standard IX
Mathematics
Property 6
Evaluate:iv ...
Question
Evaluate:
(iv)
2
5
×
15
0
+
(
−
3
)
3
−
(
2
7
)
−
2
.
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Solution
Given ,
2
5
×
15
0
+
(
−
3
)
3
−
(
2
7
)
−
2
.
On further simplification ,we have:
2
5
×
15
0
+
(
−
3
)
3
−
(
2
7
)
−
2
=
2
×
2
×
2
×
2
×
2
×
1
+
(
−
3
)
×
(
−
3
)
×
(
−
3
)
−
(
7
2
)
×
7
2
[
∵
a
0
=
1
,
a
−
m
=
1
a
m
]
=
32
×
1
−
27
−
49
4
=
32
×
4
1
×
4
−
27
×
4
1
×
4
−
49
4
×
1
(
∵
LCM
=
4
)
=
128
−
108
−
49
4
=
−
29
4
=
−
7
1
4
Hence,
2
5
×
15
0
+
(
−
3
)
3
−
(
2
7
)
−
2
=
−
7
1
4
.
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