Question

a^2/(b+c)=b^2/(c+a)=c^2/(a+b)দেখাই দেখাই যে 1/(1+a)+1/(1+b)+1/(1+c)=1​

Answer 1

Answer:

If a^2/b+c=b^2/c+a=c^2/a+b=1 then what is the value of 1/1+a + 1/1+b + 1/1+c?

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3 Answers



Shivani Prajapati

, Have great interest in mathematics

Answered 2 years ago · Author has 111 answers and 177.6K answer views

In this question

a^2=b+c , b^2=c+a , c^2=a+b is given

Now we are asked to find the value of

1/1+a + 1/1+b + 1/1+c

So to solve this we need the value of 1+a , 1+b , 1+c and to find it what we will do is add a , b and c to the given equations……. okk let’s do it

a^2=b+c

a^2+a = a+b+c

a (a+1)=a+b+c

a+1= (a+b+c)/a

For 1+b also follow same process adding b on both sides and we get

1+b= (a+b+c)/ b

and for 1+c , adding c on both sides , we get

1+c=(a+b+c)/c

Now put these values in the question whose value is to be found

1/1+a + 1/1+b + 1/1+c

a/(a+b+c) + b/(a+b+c) + c/(a+b+c)

Taking LCM :

(a+b+c )/ (a+b+c ) =1

And hence we have solved it , these questions require the need of tricks as we cannot solve it as a whole as it would be quite lengthy process hence we added a ,b and c in the beginning only so that we get something uniform out of it.

All the best ……!