Answer :

[tex]3x=4y=12z \Rightarrow x=4z,y=3z \\ \frac{xy}{x+y} \\ \\ \frac{4z3z}{4z+3z} \\ \\ \frac{12z^2}{7z} \\ \\ \frac{12}{7} z[/tex]

So, I proved that [tex]\frac{xy}{x+y} \ne z[/tex]

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