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Mohit joshi

· started a discussion

· 1 Months ago

/ (Question: 144598.0 )

3ab or 3b?

Question:
If P = \(\cfrac{3ab}{(2a+b)}\), then find the value of \(\cfrac{(P+a)}{(P-a)}\)  in terms of a and b.
Options:
A) \(\cfrac{(a+2b)}{(a-b)}\)
B) \(\cfrac{(a+2b)}{(b-a)}\)
C) \(\cfrac{(b+2a)}{(a-b)}\)
D) \(\cfrac{(b+2a)}{(b-a)}\)
Solution:

Ans (b) Given that P = \(\cfrac{3ab}{(2a+b)}\) ⇒ \(\cfrac{p}{a}\) = \(\cfrac{3ab}{(2a+b)}\)

Applying componendo – dividendo, we get

\(\cfrac{P+a}{P-a}=\cfrac{3b+2a+b}{3b-(2a+b)}\)    = \(\cfrac{4b+2a}{2b-2a}=\cfrac{2b+a}{b-a}\)


Nihit

· commented

· 1 Months ago

Yet to be approved!

Knowledge Expert

· commented

· 1 Months ago

Dear student
Given answer is correct
Please try again

Best wishes
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