Tìm x , biết: a) ( x − 2 )^2 − ( x − 3 ) ( x + 3 ) = 6 b) 2x ( x − 3 ) − 5 ( 3 − x ) = 0.

a) \({\left( {x - 2} \right)^2} - \left( {x - 3} \right)\left( {x + 3} \right) = 6\)

\({x^2} - 4x + 4 - \left( {{x^2} - 9} \right) = 6\)

\({x^2} - 4x + 4 - {x^2} + 9 = 6\)

\(\left( {{x^2} - {x^2}} \right) - 4x = 6 - 4 - 9\)

\( - 4x = - 7\)

\(x = \frac{7}{4}\)

Vậy \(x = \frac{7}{4}.\)

b) \(2x\left( {x - 3} \right) - 5\left( {3 - x} \right) = 0\)

\(2x\left( {x - 3} \right) + 5\left( {x - 3} \right) = 0\)

\(\left( {x - 3} \right)\left( {2x + 5} \right) = 0\)

Suy ra \(x - 3 = 0\) hoặc \(2x + 5 = 0\)

\(x = 3\) hoặc \(2x = - 5\)

\(x = 3\) hoặc \(x = - \frac{5}{2}.\)

Vậy \(x \in \left\{ {3; - \frac{5}{2}} \right\}.\)