1a+11a+1\frac{1}{{a + 1}} + 1b+1" role="presentation">1b+11b+1\frac{1}{{b + 1}} + 1c+1" role="presentation">1c+11c+1\frac{1}{{c + 1}} ?" />
1a+11a+1\frac{1}{{a + 1}} + 1b+1" role="presentation">1b+11b+1\frac{1}{{b + 1}} + 1c+1" role="presentation">1c+11c+1\frac{1}{{c + 1}} ?" />
1a+11a+1\frac{1}{{a + 1}} + 1b+1" role="presentation">1b+11b+1\frac{1}{{b + 1}} + 1c+1" role="presentation">1c+11c+1\frac{1}{{c + 1}} ?" />
India’s No.1 Educational Platform For UPSC,PSC And All Competitive Exam
If a2 = b + c, b2 = a + c, c2 = b + a, then what will be the value of
1a+1
+ 1b+1
+ 1c+1
?
Answer:- 1
Explanation:-
Solution:
a2=b+c, b2=a+c, c2=b+aTaking a=2, b=2 and c=2So,(2)2=2+24=4Now,1a+1+ 1b+1+ 1c+1Put a=2, b=2 and c=2=12+1+ 12+1+ 12+1=13+13+13=1
+ 1b+1
+ 1c+1
?",
"text": "If a2 = b + c, b2 = a + c, c2 = b + a, then what will be the value of 1a+1
+ 1b+1
+ 1c+1
?",
"dateCreated": "7/24/2019 10:09:12 AM",
"author": {
"@type": "Person",
"name": "Nitin Sir"
},
"answerCount": "4",
"acceptedAnswer": {
"@type": "Answer",
"text": "
Solution:
a2=b+c, b2=a+c, c2=b+aTaking a=2, b=2 and c=2So,(2)2=2+24=4Now,1a+1+ 1b+1+ 1c+1Put a=2, b=2 and c=2=12+1+ 12+1+ 12+1=13+13+13=1
",
"dateCreated": "7/24/2019 10:09:12 AM",
"author": {
"@type": "Person",
"name": "Nitin Sir"
}
},
"suggestedAnswer": {
"@type": "Answer",
"text": "
Solution:
a2=b+c, b2=a+c, c2=b+aTaking a=2, b=2 and c=2So,(2)2=2+24=4Now,1a+1+ 1b+1+ 1c+1Put a=2, b=2 and c=2=12+1+ 12+1+ 12+1=13+13+13=1
",
"dateCreated": "7/24/2019 10:09:12 AM"
}
}
Post your Comments