mateforum.ro

Sa se determine nmerele reale x si y stiind ca:

\( \left{\begin{array}{cc}x^2+(2y-3)^2=5\\xy=4\end{array} \)

Avem ca \( x^{2}+\left(2y-3\right)^{2}=5\Longleftrightarrow x^{2}+4y^{2}-12y+4=0 \) si cum \( x=\frac{4}{y} \) avem: \( y^{4}-3y^{3}+y^{2}+4=0\Longleftrightarrow\left(y-2\right)\left(y^{3}-y^{2}-y-2\right)=0\Longleftrightarrow\left(y-2\right)^{2}\left(y^{2}+y+1\right)=0. \). Cum \( y^{2}+y+1\neq0 \) rezulta \( y=2 \) si \( x=2 \).