\( x△y =\frac{ xy}{x +y}\) and we need to find the value of\(\frac{1}{a}△(\frac{1}{b}△\frac{1}{c})\) ?
\(\frac{1}{b}△\frac{1}{c}\) = (1/b)*(1/c) / (1/b + 1/c) = (1/bc) / ((b+c)/bc) = 1/(b+c)
Similarly,\(\frac{1}{a}\) △\((\frac{1}{b}△\frac{1}{c})\) =\(\frac{ 1 }{ a + b +c}\)
So, Answer will beC
Hope ithelps!
Watch the following video to learn the Basics of Functions and CustomCharacters
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\(\frac{1}{b}△\frac{1}{c}\) = (1/b)*(1/c) / (1/b + 1/c) = (1/bc) / ((b+c)/bc) = 1/(b+c)
Similarly,\(\frac{1}{a}\) △\((\frac{1}{b}△\frac{1}{c})\) =\(\frac{ 1 }{ a + b +c}\)
So, Answer will beC
Hope ithelps!
Watch the following video to learn the Basics of Functions and CustomCharacters
Iframe
...
Statistics : Posted by BrushMyQuant • on 15 Jun 2023, 06:00 • Replies 2 • Views 858
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