Răspuns:
Explicație pas cu pas:
[tex]\texttt{Se formeaza patrate perfecte:}\\\sqrt{x+2\cdot\sqrt{x-1}}+\sqrt{x-2\cdot \sqrt{x-1}}=2\\\sqrt{x-1+2\cdot\sqrt{x-1}+1}+\sqrt{x-1-2\cdot\sqrt{x-1}+1}=2\\\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}=2\\|\sqrt{x-1}+1|+|\sqrt{x-1}-1||=2\\\texttt{Conditii de existenta : }x\geq 1:\\\texttt{Daca }x\in {[1,2)}:\\\sqrt{x-1}+1+1-\sqrt{x-1}=2\\2=2,\texttt{adevarat}\\\texttt{Daca }x\in{[2,\infty)}:\\\sqrt{x-1}+1+\sqrt{x-1}-1=2\\2\sqrt{x-1}=2\\\sqrt{x-1}=1\\x-1=1\\x=2\\S={[1,2]}[/tex]
Raspunsul este c)
