[tex]\sqrt{x+1}+\sqrt{x+6} = 5\\\\ $Conditii de existenta: $ x+1\geq0 $ si $ x+6\ \textgreater \ 0 \Rightarrow x\geq -1 $ si $ x\geq -6 \Rightarrow \\ \Rightarrow D = [-1,+\infty)\\\\$Notam: $ a = \sqrt{x+1},\quad b = \sqrt{x+6} \\ \\ $Facem sistem:
\left\{ \begin{array}{c} a+b = 5\quad \quad \quad \quad \quad \quad \quad \quad \quad $ $ $ $ \\ a^2-b^2 = x+1-(x+6) = -5 \end{array} \right \Rightarrow \\ \\
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[tex]\Rightarrow \left\{ \begin{array}{c} a+b=5 \\ a^2-b^2 = -5 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a+b=5 \\ (a-b)(a+b) = -5 \end{array} \right \Rightarrow \\ \\[/tex]
[tex] \Rightarrow \left\{ \begin{array}{c} a+b=5 \\ (a-b)\cdot 5 = -5 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a+b=5 \\ (a-b) =\dfrac{-5}{5}\end{array} \right \Rightarrow \left\{ \begin{array}{c} a+b=5 \\ a-b=-1 \end{array} \right \Rightarrow \\ ~\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad------ (+) \\ \Rightarrow 2a = 4 \Rightarrow a = 2 \Rightarrow \sqrt{x+1}=2 \Rightarrow x+1 = 4 \Rightarrow \boxed{x = 3}\quad \boxed{I} \\ \\ [/tex]
[tex]a+b=5 \Rightarrow 2+b=5 \Rightarrow b = 5-2 \Rightarrow b = 3 \Rightarrow \sqrt{x+6} = 3 \Rightarrow \\ \\ \Rightarrow x+6 = 9 \Rightarrow x = 9-6 \Rightarrow \boxed{x = 3} \quad \boxed{II} \\ \\ $Din \boxed{I}$ $ $\cup$ $\boxed{II} \Rightarrow \boxed{S = \Big\{3\Big\}}[/tex]
