Question

sa se determine solutiile reale ale ecuatiei:
√(x^2-4)+√(x-2)=0

Answer 1

Explicație pas cu pas:

√(x^2-4)+√(x-2)=0

DVA: x∈(-∞,-2]∪[2,+∞)

√[(x-2)(x+2)]+√(x-2)=0

√(x-2)(√(x+2)+1)=0

√(x-2)=0⇒x=2∈DVA

√(x+2)+1=0⇒√(x+2)=-1⇒x∈∅

Raspuns: S={2}

Answer 2

$\sqrt{x^2-4}+\sqrt{x-1} = 0 \\ \\ \begin{cases}\sqrt{x^2-4}\geq 0\\\sqrt{x-1}\geq 0 \end{cases} \Rightarrow \begin{cases} \sqrt{x^2-4} = 0\\\sqrt{x-2} = 0 \end{cases}\Rightarrow \begin{cases} x^2-4 = 0\\x-2 = 0 \end{cases}\Rightarrow \\ \\ \Rightarrow \begin{cases} x^2 = 4\\ x = 2 \end{cases}\Rightarrow \begin{cases} x = \pm 2\\x=2 \end{cases}\\ \\\\ \Rightarrow S=\in\{-2,+2\} \cap\{2\} \Rightarrow \boxed{S= \{2\}}$