Chủ đề này có 2 trả lời

[trang2004]

$8x^2+\frac{1}{x}=\frac{5}{2}$

[DOTOANNANG]

Đặt $4\,x= \frac{s}{3}+ \frac{5}{s}\,\left ( s\neq 0 \right )$, ta có: $0= 8\,x^{2}+ \frac{1}{x}- \frac{5}{2}= \frac{\left ( 4\,x \right )^{3}- 5\left ( 4\,x \right )+ 8}{2\left ( 4\,x \right )}= \left [ \left ( \frac{s}{3} \right )^{3}+ \left ( \frac{5}{s} \right )^{3}+ 8 \right ]\,\left ( 8\,x \right )^{-\,1}$

nên $s^{3}$ là nghiệm của phương trình: ${s_{\,0}}^{\,2}+ 216\,s_{\,0}+ 3\,375= 0$

$$\Leftrightarrow {s_{\,1,\,2}}^{\,\,3}= \pm 3\,\sqrt{921}- 108\Leftrightarrow s_{\,1,\,2}= \sqrt[3]{\pm 3\,\sqrt{921}- 108}$$

nên:

$x= \frac{1}{4}\left ( \frac{s_{\,1}}{3}+ \frac{5}{s_{\,1}} \right )= \frac{1}{4}\left ( \frac{s_{\,2}}{3}+ \frac{5}{s_{\,2}} \right )= \frac{1}{4}\left ( -\,\frac{5}{\sqrt[3]{108- 3\sqrt{921}}}- \frac{\sqrt[3]{36- \sqrt{921}}}{3^{\,2\,/\,3}} \right )$

 

[toanhoc2017]

Đặt $4\,x= \frac{s}{3}+ \frac{5}{s}\,\left ( s\neq 0 \right )$, ta có: $0= 8\,x^{2}+ \frac{1}{x}- \frac{5}{2}= \frac{\left ( 4\,x \right )^{3}- 5\left ( 4\,x \right )+ 8}{2\left ( 4\,x \right )}= \left [ \left ( \frac{s}{3} \right )^{3}+ \left ( \frac{5}{s} \right )^{3}+ 8 \right ]\,\left ( 8\,x \right )^{-\,1}$
nên $s^{3}$ là nghiệm của phương trình: ${s_{\,0}}^{\,2}+ 216\,s_{\,0}+ 3\,375= 0$
$$\Leftrightarrow {s_{\,1,\,2}}^{\,\,3}= \pm 3\,\sqrt{921}- 108\Leftrightarrow s_{\,1,\,2}= \sqrt[3]{\pm 3\,\sqrt{921}- 108}$$
nên:
$x= \frac{1}{4}\left ( \frac{s_{\,1}}{3}+ \frac{5}{s_{\,1}} \right )= \frac{1}{4}\left ( \frac{s_{\,2}}{3}+ \frac{5}{s_{\,2}} \right )= \frac{1}{4}\left ( -\,\frac{5}{\sqrt[3]{108- 3\sqrt{921}}}- \frac{\sqrt[3]{36- \sqrt{921}}}{3^{\,2\,/\,3}} \right )$