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Grade: 12th pass

                        

If a+b+c=3 then what is the maximum valuethe equation:a/(a^3+b^2+c) + b/(b^3+c^2+a) + c/(c^3+a^2+b)

one month ago

Answers : (1)

NANDEESH
13 Points
							
16t²–40t–119=0
Write 119 as a factor of its prime factors.
 
119=7*17
 
Now,
See if we can find two numbers such that:
Product=16*(–119)
And,
Sum=–40
 
Thus, the numbers are: 28 and–68
 
Therefore, the equation becomes:
=〉16t²–68t+28t–119=0
=〉 4t(4t–17)–7(4t–17)=0
=〉(4t–7)(4t–17)=0
 
=〉t = 7/4 or 17/4
Thus,7/4 and17/4 are the required solutions!
one month ago
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