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Sa se gaseasca perechile de numere reale (n,m) , m>0 pentru care ecuatia

\( \frac{x}{\sqrt{m}+\frac{x}{\sqrt{m}+\frac{x}{\sqrt{m}}}}+(1-n)\sqrt{m}=0 \)

admite o singura radacina reala. Exista perechi (n,m), \( n,m\in \mathb{R}, m>0 \) pentru care aceasta ecuatie nu admite solutii reale?

M.Popescu, ,,Gh.Titeica'' 2005

Deoarece \( \frac{x}{\sqrt{m}+\frac{x}{m+\frac{x}{\sqrt{m}}}}=\frac{x}{\sqrt{m}+\frac{x\sqrt{m}}{m+x}}=\frac{x\left(m+x\right)}{\sqrt{m}\left(m+2x\right)} \) ecuatia devine \( \frac{x\left(m+x\right)}{\sqrt{m}\left(m+2x\right)}+\left(1-n\right)\sqrt{m}=0\Longleftrightarrow\frac{x\left(m+x\right)}{m+2x}+m\left(1-n\right)=0\Longleftrightarrow x+m-mn=\frac{x^{2}}{m+2x} \) adica \( \left(m+2x\right)\left(x+m-mn\right)=x^{2}\Longleftrightarrow x^{2}+xm\left(3-2n\right)+m^{2}-m^{2}n=0 \). Ecuatia are 2 radacini reale distincte, deoarece \( \Delta>0\Longleftrightarrow m^{2}\left(3-2n\right)^{2}>4m^{2}\left(1-n\right)\Longleftrightarrow\left(3-2n\right)^{2}>4-4n\Longleftrightarrow4n^{2}-8n+5>0\Longleftrightarrow4\left(n-1\right)^{2}+1>0 \).

Ai grija de conditiile de existenta