Đáp án:

Giải thích các bước giải:

$\sqrt[3]{x + \frac{1}{2}} = 16x³ - 1$

$⇔\sqrt[3]{x + \frac{1}{2}} - 1 - 16(x³ - \frac{1}{8}) = 0$

$⇔(\sqrt[3]{x + \frac{1}{2}})³ - 1³ - 16(x³ - \frac{1}{8})[(\sqrt[3]{x + \frac{1}{2}})² + \sqrt[3]{x + \frac{1}{2}} + 1] = 0 $

$⇔(x - \frac{1}{2}) - 16(x - \frac{1}{2})(x² + \frac{x}{2} + \frac{1}{4})[(\sqrt[3]{x + \frac{1}{2}})² + \sqrt[3]{x + \frac{1}{2}} + 1] = 0$

$⇔(x - \frac{1}{2})[1 - 16(x² + \frac{x}{2} + \frac{1}{4})[(\sqrt[3]{x + \frac{1}{2}})² + \sqrt[3]{x + \frac{1}{2}} + 1] = 0$

@ $ x - \frac{1}{2} = 0 ⇔ x =\frac{1}{2}$

@ $ 1 - 16(x² + \frac{x}{2} + \frac{1}{4})[(\sqrt[3]{x + \frac{1}{2}})² + \sqrt[3]{x + \frac{1}{2}} + 1] = 0$

$ ⇔ 16(x² + \frac{x}{2} + \frac{1}{4})[(\sqrt[3]{x + \frac{1}{2}})² + \sqrt[3]{x + \frac{1}{2}} + 1] = 1$ Vô nghiệm vì :

$ VT = 16[(x + \frac{1}{4})²+ \frac{3}{16}].[(\sqrt[3]{x + \frac{1}{2}} + \frac{1}{2})² + \frac{3}{4}] ≥ 16.\frac{3}{16}.\frac{3}{4} = \frac{9}{4} > 1$

Vậy $PT$ có nghiệm duy nhất $ x =\frac{1}{2}$