[tex] \sqrt{5+x} [/tex] este definit pentru x≥-5.
[tex] \sqrt{5-x} \geq 0=\ \textgreater \ x-1 \geq 0=\ \textgreater \ \boxed{x \geq 1}.[/tex]
Ridicam ecuatia la patrat:
=> [tex] x^{2} -2x+1=5+x\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ x^{2} -3x-4=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ x^{2} -3x-3-1=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ ( x^{2} -1)-3(x+1)=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ (x+1)(x-1)-3(x+1)=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ (x+1)(x-4)=0=\ \textgreater \ x=-1~sau~x=4.[/tex]
x=-1 nu convine pentru ca -1<1.
Deci unica solutie este: [tex]\boxed{x=4}.[/tex]
