Răspuns:
Explicație pas cu pas:
[tex]\texttt{Conditii de existenta }x\in[0,\infty)\\\sqrt{2+x}\cdot 1-1\cdot x=0\\\sqrt{2+x}=x|()^2\\x+2=x^2\\x^2-x-2=0\\\Delta=1+8=9\Rightarrow \sqrt{\Delta}=3\\x_1=\dfrac{1+3}{2}=\dfrac{4}{2}=2\\\\x_2=\dfrac{1-3}{2}=-\dfrac{2}{2}=-1\texttt{ nu convine}\\S=2[/tex]
