Đáp án đúng: B
Giải chi tiết:\(\sqrt {3{x^2} - 7x + 3} - \sqrt {{x^2} - 2} = \sqrt {3{x^2} - 5x - 1} - \sqrt {{x^2} - 3x + 4} \)
Điều kiện : \(\left\{ \begin{array}{l}3{x^2} - 7x + 3 \ge 0\\{x^2} - 2 \ge 0\\3{x^2} - 5x - 1 \ge 0\end{array} \right.(*)\)
Ta có :
\(\begin{array}{l}\sqrt {3{x^2} - 7x + 3} - \sqrt {{x^2} - 2} = \sqrt {3{x^2} - 5x - 1} - \sqrt {{x^2} - 3x + 4} \\ \Leftrightarrow \sqrt {3{x^2} - 7x + 3} - \sqrt {3{x^2} - 5x - 1} + \sqrt {{x^2} - 3x + 4} - \sqrt {{x^2} - 2} = 0\\ \Leftrightarrow \dfrac{{\left( {\sqrt {3{x^2} - 7x + 3} - \sqrt {3{x^2} - 5x - 1} } \right).\left( {\sqrt {3{x^2} - 7x + 3} + \sqrt {3{x^2} - 5x - 1} } \right)}}{{\sqrt {3{x^2} - 7x + 3} + \sqrt {3{x^2} - 5x - 1} }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \dfrac{{\left( {\sqrt {{x^2} - 3x + 4} - \sqrt {{x^2} - 2} } \right)\left( {\sqrt {{x^2} - 3x + 4} + \sqrt {{x^2} - 2} } \right)}}{{\sqrt {{x^2} - 3x + 4} + \sqrt {{x^2} - 2} }} = 0\\ \Leftrightarrow \left( { - 2x + 4} \right).\dfrac{1}{{\sqrt {3{x^2} - 7x + 3} + \sqrt {3{x^2} - 5x - 1} }} + \left( { - 3x + 6} \right).\dfrac{1}{{\sqrt {{x^2} - 3x + 4} + \sqrt {{x^2} - 2} }} = 0\\ \Leftrightarrow \left( { - x + 2} \right)\underbrace {\left( {\dfrac{2}{{\sqrt {3{x^2} - 7x + 3} + \sqrt {3{x^2} - 5x - 1} }} + \dfrac{3}{{\sqrt {{x^2} - 3x + 4} + \sqrt {{x^2} - 2} }}} \right)}_{ > 0} = 0\\ \Leftrightarrow - x + 2 = 0 \Leftrightarrow x = 2\,\,\left( {tm\,\,\left( * \right)} \right)\end{array}\)
Vây \(x = 2\) là nghiệm duy nhất của phương trình.
