`x^4-6x^3+8x^2+2x-1=0`

Đáp án+Giải thích các bước giải:

\(x^4 - 6x^3 + 8x^2 + 2x -1=0\\\Leftrightarrow x^4 - 4x^3 - 2x^3 + x^2 + 8x^2 - x^2 - 2x +4x -1=0\\\Leftrightarrow x^2(x^2 - 4x + 1)-2x(x^2 - 4x +1)-(x^2 - 4x+1)=0\\\Leftrightarrow (x^2 - 4x +1)(x^2 - 2x -1)=0 \\\Leftrightarrow\left[ \begin{array}{l}x^2 -4x +1=0\\\\x^2 -2x -1=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{-(-4)\pm\sqrt{(-4)^2 - 4\times 1\times 1}}{2\times 1}\\\\x=\dfrac{-(-2)\pm \sqrt{(-2)^2 -4\times 1\times (-1)}}{2\times 1}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{4\pm\sqrt{12}}{2}\\\\x=\dfrac{2\pm\sqrt{8}}{2}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{4\pm 2\sqrt{3}}{2}\\\\x=\dfrac{2\pm2\sqrt{2}}{2}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x=\dfrac{4+2\sqrt{3}}{2}\\x=\dfrac{4-2\sqrt{3}}{2}\end{array} \right.\\\left[ \begin{array}{l}x=\dfrac{2+2\sqrt{2}}{2}\\x=\dfrac{2-2\sqrt{2}}{2}\end{array} \right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=2+\sqrt{3}\\x=2-\sqrt{3}\\x=1+\sqrt{2}\\x=1-\sqrt{2}\end{array} \right.\)