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Concursul "Teodor Topan" - problema 2
Posted: Mon Dec 03, 2007 2:30 pm
by maky
Sa se afle \( x,y,z \) stiind ca: \( \frac{x}{1}=\frac{y^2}{8}=\frac{z^3}{4} \)si \( x\cdot{y}\cdot{z}=16 \)
Radu Florica, Simleu Silvaniei
Posted: Mon Mar 03, 2008 5:11 pm
by Marius Dragoi
Observam ca \( x \) si \( z \) sunt pozitive.
\( z=2^{\frac{2}{3}} x^{\frac{1}{3}} \) , \( y=2^{\frac{3}{2}} x^{\frac{1}{2}} \)
Din \( x y z=16 \) si relatiile precedente obtinem: \( 2^{\frac{2}{3}+\frac{3}{2}} x ^{1+\frac{1}{2}+\frac{1}{3}}=16 \)
\( \Rightarrow \) \( x=2 \) \( \Rightarrow \) \( z=2 \) si \( y=4 \)