Question

If abc=1 , prove \dfrac{a}{1+a+ab}+\dfrac{b}{1+b+bc}+\dfrac{c}{1+c+ca} =1

Collected in the board: Fractions

Steven Zheng posted 1 year ago


Answer

\because abc = 1,

\dfrac{a}{1+a+ab} =\dfrac{a}{abc+a+ab} = \dfrac{1}{1+b+bc}

\therefore \dfrac{a}{1+a+ab}+\dfrac{b}{1+b+bc}+\dfrac{c}{1+c+ca}

= \dfrac{1}{1+b+bc}+\dfrac{b}{1+b+bc}+\dfrac{c}{1+c+ca}

=\dfrac{1+b}{1+b+bc} +\dfrac{c}{1+c+ca}

=\dfrac{1+b+bc-bc}{1+b+bc} +\dfrac{c}{1+c+ca}

=1-\dfrac{bc}{1+b+bc} +\dfrac{c}{1+c+ca}

=1-\dfrac{bc}{abc+b+bc} +\dfrac{c}{1+c+ca}

=1

Steven Zheng posted 1 month ago

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