Question

Să se rezolve în mulţimea numerelor reale ecuaţia $\sqrt{2-x} + \sqrt[3]{x-2} = 0$

Answer 1

Rezolvarea se afla in fotografia de mai jos

Answer 2

$\\ Conditii de existenta: 2-x \geq 0 \Rightarrow -x \geq -2 \Rightarrow x\leq 2 \Rightarrow D = (-\infty, 2]$

$\sqrt{2-x} + \sqrt[3]{x-2}= 0 \Rightarrow \sqrt{2-x} + \sqrt[3]{-(2-x)} = 0 \Rightarrow \\ \\ \Rightarrow \sqrt{2-x} - \sqrt[3]{2-x} = 0 \Rightarrow (2-x)^{\dfrac{1}{2}} - (2-x)^{\dfrac{1}{3}} = 0 \Rightarrow \\ \\ \Rightarrow (2-x)^{\dfrac{1}{3}}\cdot \Big((2-x)^{\dfrac{1}{2}-\dfrac{1}{3}}-1\Big) = 0 \Rightarrow \\ \\ \Rightarrow (2-x)^{\dfrac{1}{3}}\cdot \Big((2-x)^{{\dfrac{3-2}{6}} }- 1\Big) = 0 \Rightarrow$

$\Rightarrow (2-x)^{\dfrac{1}{3}}\cdot \Big((2-x)^{{\dfrac{1}{6}} }- 1\Big) = 0 \Rightarrow \sqrt[3]{2-x} \cdot \Big(\sqrt[6]{2-x}-1\Big) = 0 \\ \\ \boxed{1} \quad \sqrt[3]{2-x} = 0 \Rightarrow 2-x = 0 \Rightarrow \boxed{x = 2} \\ \\ \boxed{2} \quad \sqrt[6]{2-x}-1 = 0 \Rightarrow \sqrt[6]{2-x} = 1 \Rightarrow 2-x = 1 \Rightarrow -x = -1 \Rightarrow \\ \\ \Rightarrow \boxed{x=1} \\ \\ Din \Big(\boxed{1} \cup \boxed{2}\Big) \cap D \quad \Rightarrow \boxed{x \in \Big\{1,2\Big\}}$