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Standard VI
Maths
Divisibility Rules
Question
Prove that $$\dfrac {1}{9!}+\dfrac {1}{10!}+\dfrac {1}{11!}=\dfrac {122}{11!}$$
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Solution
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Similar Questions
Q1
Prove that:
1
9
!
+
1
10
!
+
1
11
!
=
122
11
!
View Solution
Q2
Prove that:
1
9
!
+
1
10
!
+
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11
!
=
122
11
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Q3
If
1
√
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−
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√
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+
1
√
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×
10
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√
99
x
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then x equals
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Q4
Prove that :
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−
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1
4
+
tan
−
1
2
9
=
1
2
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−
1
4
5
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Q5
Prove that:
tan
−
1
1
4
+
tan
−
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2
9
=
1
2
cos
−
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3
5
.
View Solution