Tìm x, biết: \(3\sqrt {x - 2} - \sqrt {{x^2} - 4} = 0.\)

ĐK: x ≥ 2.

\(3\sqrt {x - 2} - \sqrt {{x^2} - 4} = 0\)

⇔ \(3\sqrt {x - 2} - \sqrt {\left( {x + 2} \right)\left( {x - 2} \right)} = 0\)

⇔ \(\sqrt {x - 2} \left( {3 - \sqrt {x + 2} } \right) = 0\)

⇔ \(\left[ {\begin{array}{*{20}{c}}{\sqrt {x - 2} = 0}\\{3 - \sqrt {x + 2} = 0}\end{array}} \right.\)

⇔ \(\left[ {\begin{array}{*{20}{c}}{x = 2\,\,\left( {tm} \right)}\\{x = 7\,\,\left( {tm} \right)}\end{array}} \right.\)

Vậy x = 2 hoặc x = 7.