Question

5x⁴-8x²+1=0
помогите пожалуйста ​

Answer 1

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$\displaystyle \bf 5x^4-8x^2+1=0$

$\displaystyle \bf x^2 = t$            $\boxed{\displaysyle \bf t > 0}$

$\displastyle \bf \\\\ 5t^2 -8t +1 =0$

$\displastylr\bf D =b^2-4ac=(-8)^2 -4\cdot 5\cdot 1 =44$

$\displatyle \bf \sqrt{D} = \pm \sqrt{44}=\pm2\sqrt{11}$

$\displastyle \bf t = \frac{1}{10}\cdot \Big(8\pm2\sqrt{11} \Big)$

$\displaystyle\bf t_1= \frac{8 - 2\sqrt{11}}{10} = \frac{4-\sqrt11}{5} \ \Rightarrow \ x^2 = \frac{4-\sqrt{11}}{5}$

$\boxed{\displaystyle\bf X_1 = -\sqrt{\frac{4-\sqrt{11}}{5}}\\\\}$          $\boxed{\displaystyle\bf X_2 = \sqrt{\frac{4-\sqrt{11}}{5}}}$

$\displaystyle \bf t_2 = \frac{8 + 2\sqrt{11}}{10} = \frac{4+\sqrt{11}}{5}\ \Rightarrow \ x^2 =\frac{4+\sqrt{11}}{5}$

$\boxed{\displaystyle \bf X_3 = - \sqrt{\frac{4+\sqrt{11}}{5}}}$           $\boxed{\displaystyle \bf X_4 = \sqrt{\frac{4+\sqrt{11}}{5}}}$  

Answer 2

Ответ:

$x\in\{\frac{\sqrt{20+5\sqrt{11}}}{5}}; -\frac{\sqrt{20+5\sqrt{11}}}{5}};\frac{\sqrt{20-5\sqrt{11}}}{5}};-\frac{\sqrt{20-5\sqrt{11}}}{5}}\}$

Объяснение:

$5x^{4}-8x^{2}+1=0\\x^{2}=t\\5t^{2}-8t+1=0\\a=5;b =-8;c=1\\\Delta=\sqrt{b^{2}-4\cdot a\cdot c}\\\Delta=\sqrt{(-8)^{2}-4\cdot5\cdot1}\\\Delta=\sqrt{64-20\cdot1}\\\Delta=\sqrt{64-20}\\\Delta=\sqrt{44}\\\Delta=\sqrt{4\cdot11}\\\Delta=\sqrt{4}\cdot\sqrt{11}\\\Delta=2\sqrt{11}\\t_{1,2}=\frac{-b\pm\Delta}{2\cdot a}\\t_{1,2}=\frac{-(-8)\pm2\sqrt{11}}{2\cdot5}\\t_{1,2}=\frac{8\pm2\sqrt{11}}{10}\\t_1=\frac{8+2\sqrt{11}}{10}\ \ \ t_2=\frac{8-2\sqrt{11}}{10}$
$t_1=\frac{4+\sqrt{11}}{5}\ \ \ \ t_2=\frac{4-\sqrt{11}}{5}\\x^{2}=\frac{4+\sqrt{11}}{5}\ \ \ \ \ \ \ \ \ \ \ x^{2}=\frac{4-\sqrt{11}}{5}\\x=\pm\sqrt{\frac{4+\sqrt{11}}{5}}\ \ \ \ \ \ \ x=\pm\sqrt{\frac{4-\sqrt{11}}{5}}\\x=\pm\frac{\sqrt{4+\sqrt{11}}}{\sqrt{5}}\ \ \ \ \ \ \ \ x=\pm\frac{\sqrt{4-\sqrt{11}}}{\sqrt{5}}}\\x=\pm\frac{\sqrt{5}\cdot\sqrt{4+\sqrt{11}}}{\sqrt{5}\cdot\sqrt{5}}\ \ \ \ x=\pm\frac{\sqrt{5}\cdot\sqrt{4-\sqrt{11}}}{\sqrt{5}\cdot\sqrt{5}}}$
$x=\pm\frac{\sqrt{20+5\sqrt{11}}}{5}}\ \ \ \ x=\pm\frac{\sqrt{20-5\sqrt{11}}}{5}}\\x\in\{\frac{\sqrt{20+5\sqrt{11}}}{5}}; -\frac{\sqrt{20+5\sqrt{11}}}{5}};\frac{\sqrt{20-5\sqrt{11}}}{5}};-\frac{\sqrt{20-5\sqrt{11}}}{5}}\}$