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Conc.interj."Grigore Moisil" Urziceni 2010 probl.2
Posted: Tue Mar 23, 2010 8:25 pm
by Andi Brojbeanu
Aratati ca daca \( x, y\in \mathb{R}^+ \) si \( 5x^3+2x^2y=2y^3+5xy^2 \), atunci \( \frac{3x+4y}{4x+3y}\in \mathb{Z} \).
Gazeta Matematica
Posted: Mon May 24, 2010 12:19 pm
by Andi Brojbeanu
\( x^2(5x+2y)=y^2(2y+5x) \).
Daca \( 5x+2y\neq 0 \),\( x^2=y^2\Rightarrow x^2=y^2\Rightarrow x=y \) sau \( x=-y \), deci fractia \( \frac{3x+4y}{4x+3y} \) este egala cu \( \frac{7x}{7x}=1\in \mathb{Z} \) sau cu \( \frac{-x}{x}=-1\in \mathb {Z} \).
Daca \( 5x+2y=0\Rightarrow y=-\frac{5x}{2} \), deci fractia \( \frac{3x+4y}{4x+3y} \) este egala cu \( \frac{3x-10y}{\frac{8x-15x}{2}}=\frac{-14x}{7x}=-2\in \mathb{Z} \).
Posted: Thu Jun 03, 2010 8:16 pm
by moldovan ana
altfel: deoarece expresiile sunt omogene se face substitutia x = ty si asa devine mult mai simpla.