Questão 1601245

\(\sqrt{x}\cdot\sqrt{y}=\sqrt{x\cdot y}\)

\(\sqrt{x}+\sqrt{y}=\sqrt{x+y}\)

\(\left(\sqrt{x}+\sqrt{y}\right)^2=x+y\)

\(\sqrt{x}-\sqrt{y}=\sqrt{x-y}\)