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} $$