tìm x: x+ căn(x-2)=2 nhân căn(x-1)

Đáp án:

\(x=2\) Lời giải: \(\eqalign{ & x + \sqrt {x - 2} = 2\sqrt {x - 1} \,\,\left( {x \ge 2} \right) \cr & Dat\,\,\sqrt {x - 1} = t\,\,\left( {t \ge 1} \right) \Leftrightarrow x - 1 = {t^2} \cr & \Leftrightarrow \left\{ \matrix{ x = {t^2} + 1 \hfill \cr x - 2 = {t^2} - 1 \hfill \cr} \right. \cr & PT:\,\,{t^2} + 1 + \sqrt {{t^2} - 1} = 2t \cr & \Leftrightarrow {t^2} - 2t + 1 + \sqrt {{t^2} - 1} = 0 \cr & \Leftrightarrow {\left( {t - 1} \right)^2} + \sqrt {{t^2} - 1} = 0 \cr & Ta\,\,co: \cr & \left\{ \matrix{ {\left( {t - 1} \right)^2} \ge 0 \hfill \cr \sqrt {{t^2} - 1} \ge 0 \hfill \cr} \right. \Rightarrow {\left( {t - 1} \right)^2} + \sqrt {{t^2} - 1} \ge 0 \cr & Dau\,\, = \,\,xay\,\,ra \cr & \Leftrightarrow \left\{ \matrix{ t - 1 = 0 \hfill \cr {t^2} - 1 = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ t = 1 \hfill \cr t = \pm 1 \hfill \cr} \right. \Leftrightarrow t = 1\,\,\left( {tm} \right) \cr & \Rightarrow \sqrt {x - 1} = 1 \Leftrightarrow x - 1 = 1 \Leftrightarrow x = 2\,\,\left( {tm} \right) \cr} \)