Question

Rezolvati in multimea numerelor reale ecuatia

√x-2 = ∛x-2

Answer 1

$\sqrt{x-2} = \sqrt[\big 3]{x-2} \\ (x-2)^{\dfrac{1}{2} }= (x-2)^{\dfrac{1}{3}} \quad (\text{aducem puterile la acelasi numitor}) \\ (x-2)^{\dfrac{3}{6}} = (x-2)^{\dfrac{2}{6}} \\ (x-2)^{\big 3\cdot \dfrac{1}{6}} = (x-2)^{\big 2\cdot \dfrac{1}{6}}\\ \Big((x-2)^{\big 3}\Big)^{\dfrac{1}{6}} = \Big((x-2)^{\big 2}\Big)^{\dfrac{1}{6}} \\ \\ (x-2)^3 = (x-2)^2 \\ \\ (x-2)^3-(x-2)^2 = 0 \\ \\ (x-2)^2\cdot (x-2) - (x-2)^2 = 0\\ \\ (x-2)^2\cdot \Big((x-2)-1\Big) = 0 \\ \\ (x-2)^2\cdot (x-3) = 0 \\ \\ \bullet \quad (x-2)^2 = 0 \Rightarrow x = 2 \\ \\ \bullet \quad (x-3) = 0 \Rightarrow x = 3 \\ \\ \Rightarrow \boxed{S = \Big\{2;3\Big\}}$

Answer 2

conditii de existenta , x-2≥0⇒ x≥2
fie x-2=t
atunci √t=∛t
ridicam la puterea a 6-a (putere para, vor fi posibil introduse solutii in plus, trebuie verificate)
t³=t²
t³-t²=0
t²(t-1)=0
t1=0...x-2=0...x1=2∈Domeniuluide existeanta si verifica √0=∛0
t2=1...x-2=1 ...x2=3∈Domeniului de existenta si verifica √1=∛1