Giải phương trình: $\sqrt[3]{x+\dfrac{1}{2}}$ = `16x^3-1` 24/08/2021 Bởi Sadie Giải phương trình: $\sqrt[3]{x+\dfrac{1}{2}}$ = `16x^3-1`
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}$ Bình luận
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}$