Rezolvati ecuatia:
\( \sqrt{x-y+3}+\sqrt{3y-2x+4}+\sqrt{x-2y+5}=6 \), \( x,y\in \mathbb{R} \).
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Ecuatie cu radicali
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Marius Mainea
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Solutia 1:
Din CBS
\( 36=(\sqrt{x-y+3}+\sqrt{3y-2x+4}+\sqrt{x-2y+5})^2\le(x-y+3+3y-2x+4+x-2y+5)(1+1+1)=36 \)
Asadar avem egalitate de unde \( \frac{x-y+3}{1}=\frac{3y-2x+4}{1}=\frac{x-2y+5}{1} \)
Raspuns: x=3 ,y=2;
Solutia 2: Ecuatie se scrie \( (\sqrt{x-y+3}-2)^2+(\sqrt{3y-2x+4}-2)^2+(\sqrt{x-2y+5}-2)^2=0 \)
Din CBS
\( 36=(\sqrt{x-y+3}+\sqrt{3y-2x+4}+\sqrt{x-2y+5})^2\le(x-y+3+3y-2x+4+x-2y+5)(1+1+1)=36 \)
Asadar avem egalitate de unde \( \frac{x-y+3}{1}=\frac{3y-2x+4}{1}=\frac{x-2y+5}{1} \)
Raspuns: x=3 ,y=2;
Solutia 2: Ecuatie se scrie \( (\sqrt{x-y+3}-2)^2+(\sqrt{3y-2x+4}-2)^2+(\sqrt{x-2y+5}-2)^2=0 \)