If abc = 1,then (11+a+b−1+11+b+c−1+11+c+a−1)=?If abc = 1,then (11+a+b−1+11+b+c−1+11+c+a−1)=?\eqalign{
& {\text{If }}abc{\text{ = 1,}} \cr
& {\text{then }}\left( {\frac{1}{{1 + a + {b^{ - 1}}}} + \frac{1}{{1 + b + {c^{ - 1}}}} + \frac{1}{{1 + c + {a^{ - 1}}}}} \right) = ? \cr} " />
If abc = 1,then (11+a+b−1+11+b+c−1+11+c+a−1)=?If abc = 1,then (11+a+b−1+11+b+c−1+11+c+a−1)=?\eqalign{
& {\text{If }}abc{\text{ = 1,}} \cr
& {\text{then }}\left( {\frac{1}{{1 + a + {b^{ - 1}}}} + \frac{1}{{1 + b + {c^{ - 1}}}} + \frac{1}{{1 + c + {a^{ - 1}}}}} \right) = ? \cr} " />
If abc = 1,then (11+a+b−1+11+b+c−1+11+c+a−1)=?If abc = 1,then (11+a+b−1+11+b+c−1+11+c+a−1)=?\eqalign{
& {\text{If }}abc{\text{ = 1,}} \cr
& {\text{then }}\left( {\frac{1}{{1 + a + {b^{ - 1}}}} + \frac{1}{{1 + b + {c^{ - 1}}}} + \frac{1}{{1 + c + {a^{ - 1}}}}} \right) = ? \cr} " />
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",
"text": "If a b c = 1, then ( 1 1 + a + b − 1 + 1 1 + b + c − 1 + 1 1 + c + a − 1 ) = ? If a b c = 1, then ( 1 1 + a + b − 1 + 1 1 + b + c − 1 + 1 1 + c + a − 1 ) = ?
",
"dateCreated": "7/24/2019 10:09:12 AM",
"author": {
"@type": "Person",
"name": "Nitin Sir"
},
"answerCount": "4",
"acceptedAnswer": {
"@type": "Answer",
"text": "
Solution:
Given expression, 1 1 + a + b − 1 + 1 1 + b + c − 1 + 1 1 + c + a − 1 = 1 1 + a + b − 1 + 1 b − 1 + 1 + b − 1 c − 1 + 1 a + a c + 1 = 1 1 + a + b − 1 + b − 1 1 + b − 1 + a + a a + b − 1 + 1 = 1 + a + b − 1 1 + a + b − 1 = 1 [ ∵ a b c = 1 ⇒ ( b c ) − 1 = a ⇒ b − 1 c − 1 = a , and a c = b − 1 ] Given expression, 1 1 + a + b − 1 + 1 1 + b + c − 1 + 1 1 + c + a − 1 = 1 1 + a + b − 1 + 1 b − 1 + 1 + b − 1 c − 1 + 1 a + a c + 1 = 1 1 + a + b − 1 + b − 1 1 + b − 1 + a + a a + b − 1 + 1 = 1 + a + b − 1 1 + a + b − 1 = 1 [ ∵ a b c = 1 ⇒ ( b c ) − 1 = a ⇒ b − 1 c − 1 = a , and a c = b − 1 ]
",
"dateCreated": "7/24/2019 10:09:12 AM",
"author": {
"@type": "Person",
"name": "Nitin Sir"
}
},
"suggestedAnswer": {
"@type": "Answer",
"text": "
Solution:
Given expression, 1 1 + a + b − 1 + 1 1 + b + c − 1 + 1 1 + c + a − 1 = 1 1 + a + b − 1 + 1 b − 1 + 1 + b − 1 c − 1 + 1 a + a c + 1 = 1 1 + a + b − 1 + b − 1 1 + b − 1 + a + a a + b − 1 + 1 = 1 + a + b − 1 1 + a + b − 1 = 1 [ ∵ a b c = 1 ⇒ ( b c ) − 1 = a ⇒ b − 1 c − 1 = a , and a c = b − 1 ] Given expression, 1 1 + a + b − 1 + 1 1 + b + c − 1 + 1 1 + c + a − 1 = 1 1 + a + b − 1 + 1 b − 1 + 1 + b − 1 c − 1 + 1 a + a c + 1 = 1 1 + a + b − 1 + b − 1 1 + b − 1 + a + a a + b − 1 + 1 = 1 + a + b − 1 1 + a + b − 1 = 1 [ ∵ a b c = 1 ⇒ ( b c ) − 1 = a ⇒ b − 1 c − 1 = a , and a c = b − 1 ]
",
"dateCreated": "7/24/2019 10:09:12 AM"
}
}
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