Giải phương trình : √(x^2-3x+2) +√(x^2-4x+3 ) = 2√(x^2-5x+4)

Đáp án:

x=1

Giải thích các bước giải:

$$\eqalign{ & \sqrt {{x^2} - 3x + 2} + \sqrt {{x^2} - 4x + 3} = 2\sqrt {{x^2} - 5x + 4} \cr & DKXD:\,\,\left\{ \matrix{ {x^2} - 3x + 2 \ge 0 \hfill \cr {x^2} - 4x + 3 \ge 0 \hfill \cr {x^2} - 5x + 4 \ge 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ \left[ \matrix{ x \ge 2 \hfill \cr x \le 1 \hfill \cr} \right. \hfill \cr \left[ \matrix{ x \ge 3 \hfill \cr x \le 1 \hfill \cr} \right. \hfill \cr \left[ \matrix{ x \ge 4 \hfill \cr x \le 1 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x \ge 4 \hfill \cr x \le 1 \hfill \cr} \right. \cr & PT \Leftrightarrow \sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} + \sqrt {\left( {x - 1} \right)\left( {x - 3} \right)} = 2\sqrt {\left( {x - 1} \right)\left( {x - 4} \right)} \cr & Th1:\,\,x \le 1 \cr & \Leftrightarrow \sqrt {1 - x} \sqrt {2 - x} + \sqrt {1 - x} \sqrt {3 - x} = 2\sqrt {1 - x} \sqrt {4 - x} \cr & \Leftrightarrow \sqrt {1 - x} \left( {\sqrt {2 - x} + \sqrt {3 - x} - 2\sqrt {4 - x} } \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ \sqrt {1 - x} = 0 \hfill \cr \sqrt {2 - x} + \sqrt {3 - x} = 2\sqrt {4 - x} \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = 1 \hfill \cr 2 - x + 3 - x + 2\sqrt {\left( {2 - x} \right)\left( {3 - x} \right)} = 4\left( {4 - x} \right)\,\,\left( * \right) \hfill \cr} \right. \cr & \left( * \right) \Leftrightarrow 2\sqrt {\left( {2 - x} \right)\left( {3 - x} \right)} = 11 - 2x \cr & \Leftrightarrow \left\{ \matrix{ 11 - 2x \ge 0 \hfill \cr 4\left( {{x^2} - 5x + 6} \right) = 4{x^2} - 44x + 121 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x \le {{11} \over 2} \hfill \cr 24x = 97 \hfill \cr} \right. \Leftrightarrow x = {{97} \over {24}}\,\,\left( {ktm\,\,x \le 1} \right) \cr & TH2:\,\,x \ge 4 \cr & \Leftrightarrow \sqrt {x - 1} \sqrt {x - 2} + \sqrt {x - 1} \sqrt {x - 3} = 2\sqrt {x - 1} \sqrt {x - 4} \cr & \Leftrightarrow \sqrt {x - 1} \left( {\sqrt {x - 2} + \sqrt {x - 3} - 2\sqrt {x - 4} } \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ \sqrt {x - 1} = 0 \hfill \cr \sqrt {x - 2} + \sqrt {x - 3} = 2\sqrt {x - 4} \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = 1 \hfill \cr x - 2 + x - 3 + 2\sqrt {\left( {x - 2} \right)\left( {x - 3} \right)} = 4\left( {x - 4} \right)\,\,\left( * \right) \hfill \cr} \right. \cr & \left( * \right) \Leftrightarrow 2\sqrt {\left( {x - 2} \right)\left( {x - 3} \right)} = 2x - 11 \cr & \Leftrightarrow \left\{ \matrix{ x \ge {{11} \over 2} \hfill \cr x = {{97} \over {24}}\,\,\left( {ktm} \right) \hfill \cr} \right. \cr & Vay\,\,x = 1 \cr} $$