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Question-152827




Question Number 152827 by mathdanisur last updated on 01/Sep/21
Answered by MJS_new last updated on 02/Sep/21
p=((xy)/(x+y)) ⇔ y=((px)/(x−p)) ⇔ x=((py)/(y−p))  ⇒ x>p∧y>p ∧ (x−p)∣(px) ∧ (y−p)∣(py)  ⇒  (1) y=x ⇒ y=x=2p  (2) y=p+1 ⇒ x=p(p+1)  (3) x=p+1 ⇒ y=p(p+1)
$${p}=\frac{{xy}}{{x}+{y}}\:\Leftrightarrow\:{y}=\frac{{px}}{{x}−{p}}\:\Leftrightarrow\:{x}=\frac{{py}}{{y}−{p}} \\ $$$$\Rightarrow\:{x}>{p}\wedge{y}>{p}\:\wedge\:\left({x}−{p}\right)\mid\left({px}\right)\:\wedge\:\left({y}−{p}\right)\mid\left({py}\right) \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:{y}={x}\:\Rightarrow\:{y}={x}=\mathrm{2}{p} \\ $$$$\left(\mathrm{2}\right)\:{y}={p}+\mathrm{1}\:\Rightarrow\:{x}={p}\left({p}+\mathrm{1}\right) \\ $$$$\left(\mathrm{3}\right)\:{x}={p}+\mathrm{1}\:\Rightarrow\:{y}={p}\left({p}+\mathrm{1}\right) \\ $$

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