Question

Rezolvati in multimea numerelor reale ecuatia √x+1-√x-1=√2 ajutatima va rog mult plssss

Answer 1

x≥1
x+1+x-1-2√(x²-1)=2
-2√(x²-1)=2-2x
√(x²-1)=x-1
x²-1=x²-2x+1
-2=-2x
x=1

Answer 2

$\sqrt{x+1}-\sqrt{x-1} = \sqrt2 \\ \\ Notam:\quad a = \sqrt{x+1};\quad b = \sqrt{x-1} \\ Facem sistem: \\ \\\left\{ \begin{array}{c} a-b = \sqrt2 \\ a^2-b^2 = x+1-(x-1) \end{array} \right \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a^2-b^2 = x+1-x+1 \end{array} \right \Rightarrow \\ \\ \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a^2-b^2 = 2 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ (a-b)(a+b) = 2 \end{array} \right \Rightarrow$

$\Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ \sqrt2\cdot (a+b) = 2 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a+b = \dfrac{2}{\sqrt2} \end{array} \right \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a+b = \dfrac{\sqrt2\cdot 2}{\sqrt2\cdot \sqrt2} \end{array} \right \Rightarrow$

$\Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a+b = \dfrac{\sqrt2\cdot 2}{2} \end{array} \right \Rightarrow\left\{ \begin{array}{c} a-b = \sqrt2 \\ a+b =\sqrt2}\end{array} \right |_{(adunam)} \Rightarrow \\ \\ \Rightarrow a+a-b+b = \sqrt2+\sqrt2 \Rightarrow 2a = 2\sqrt2 \Rightarrow a = \sqrt2 \Rightarrow \\ \\ \Rightarrow \sqrt{x+1} = \sqrt2 \Big|^2 \Rightarrow x+1 = 2 \Rightarrow \boxed{x = 1}$

$a-b = \sqrt2 \Rightarrow \sqrt2-b = \sqrt2 \Rightarrow b = 0 \Rightarrow \sqrt{x-1} = 0 \Rightarrow x-1 = 0\Rightarrow \\ \\ \Rightarrow \boxed{x=1} \\ \\ \\ \Rightarrow \boxed{S = \Big\{1\Big\}}$