x^3 - 3 căn 3x^2+7x- căn 3 = 0

Đáp án:

$\left[ \matrix{ x = \sqrt 3 \hfill \cr x = \sqrt 3 + \sqrt 2 \hfill \cr x = \sqrt 3 - \sqrt 2 \hfill \cr} \right.$

Giải thích các bước giải:

$\eqalign{ & {x^3} - 3\sqrt 3 {x^2} + 7x - \sqrt 3 = 0 \cr & \Leftrightarrow {x^3} - \sqrt 3 {x^2} - 2\sqrt 3 {x^2} + 6x + x - \sqrt 3 = 0 \cr & \Leftrightarrow {x^2}\left( {x - \sqrt 3 } \right) - 2\sqrt 3 x\left( {x - \sqrt 3 } \right) + \left( {x - \sqrt 3 } \right) = 0 \cr & \Leftrightarrow \left( {x - \sqrt 3 } \right)\left( {{x^2} - 2\sqrt 3 x + 1} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ x = \sqrt 3 \hfill \cr {x^2} - 2\sqrt 3 x + 1 = 0\,\,\left( 1 \right) \hfill \cr} \right. \cr & Giai\,\,\left( 1 \right):\,\,\Delta ' = 3 - 1 = 2 \cr & \Rightarrow \left[ \matrix{ x = \sqrt 3 + \sqrt 2 \hfill \cr x = \sqrt 3 - \sqrt 2 \hfill \cr} \right. \cr}$