Question

Sa se rezolve ecuatia : (radical de ordin 3 din 7x+1) - x=1

Answer 1

$\sqrt[3]{7x+1}-x=1 \\ \\ Notam: \sqrt[3]{7x+1} = t \Rightarrow 7x+1 = t^3 \Rightarrow 7x = t^3-1 \Rightarrow x = \dfrac{t^3-1}{7} \\ \\ \\ \Rightarrow t-\dfrac{t^3-1}{7}=1\Big|\cdot 7 \Rightarrow 7t-(t^3-1) = 7 \Rightarrow 7t-t^3+1 = 7 \Rightarrow \\ \\ \Rightarrow -t^3+7t+1-7 = 0 \Rightarrow -t^3+7t-6 =0 \Rightarrow t^3-7t+6 = 0 \Rightarrow \\ \\ \Rightarrow t^3-t-6t+6 = 0 \Rightarrow t(t^2-1)-6(t-1) = 0 \Rightarrow$
$\Rightarrow t(t-1)(t+1)-6(t-1) = 0 \Rightarrow (t-1)\Big(t(t+1)-6\Big) = 0 \Rightarrow \\ \\ \Rightarrow (t-1)(t^2+t-6) = 0 \\ \\ \bullet t=1 \Rightarrow \sqrt[3]{7x+1} = 1\Big|^3 \Rightarrow 7x+1 = 1 \Rightarrow 7x = 0 \Rightarrow \boxed{x = 0} \\ \\ \bullet t^2+t-6 = 0\Big| \Delta = 1+24 = 25 \Rightarrow t_{1,2} = \dfrac{-1\pm 5}{2} \Rightarrow t_1 = -3, t_2 = 2 \\ \\ t = -3 \Rightarrow \sqrt[3]{7x+1} = -3 \Big|^3 \Rightarrow 7x+1 = -27 \Rightarrow 7x = -28 \Rightarrow$

$\Rightarrow x = -\dfrac{28}{7} \Rightarrow \boxed{x = -4} \\ \\ t = 2 \Rightarrow \sqrt[3]{7x+1}=2 \Big|^3 \Rightarrow 7x+1 = 8 \Rightarrow 7x = 7 \Rightarrow x = \dfrac{7}{7} \Rightarrow \boxed{x=1}\\ \\ \\ \Rightarrow \boxed{S = \big\{-4;0;1\Big\}}$