Tìm $x$ biết $\sqrt[3]{{2x + 1}} >  - 3$.

$25a$

$5a$

$- 25{a^3}$

$- 5a$

$\dfrac{1}{{3a}}$

$\dfrac{1}{{4a}}$ 

$- \dfrac{1}{{3a}}$

$- \dfrac{1}{{8a}}$

$M > N$

$M < N$

$M \ge N$

$M + N = 0$

$\sqrt[3]{a} = 2x \Leftrightarrow {a^3} = 2x$

$\sqrt[3]{a} = 2x \Leftrightarrow 2a = {x^3}$

$\sqrt[3]{a} = 2x \Leftrightarrow a = 2{x^3}$       

$\sqrt[3]{a} = 2x \Leftrightarrow a = 8{x^3}$

$a > b \Leftrightarrow \sqrt[3]{a} > \sqrt[3]{b}$

$a < b \Leftrightarrow \sqrt[3]{a} < \sqrt[3]{b}$

$a \ge b \Leftrightarrow \sqrt[3]{a} \ge \sqrt[3]{b}$

$a < b \Leftrightarrow \sqrt[3]{a} > \sqrt[3]{b}$

$\sqrt[3]{{ab}} = \sqrt a .\sqrt b$

$\dfrac{{\sqrt[3]{a}}}{{\sqrt[3]{b}}} = \dfrac{a}{b}$ với $b \ne 0$

${\left( {\sqrt[3]{a}} \right)^3} =  - a\, khi\,a < 0$

$\dfrac{{\sqrt[3]{a}}}{{\sqrt[3]{b}}} = \sqrt[3]{{\dfrac{a}{b}}}$ với $b \ne 0$

$\sqrt[3]{{ - 125}} =  - 25$

$\sqrt[3]{{ - 125}} =  - 5$

$\sqrt[3]{{ - 125}} = 25$

$\sqrt[3]{{ - 125}} = 5$