Tìm x , biết: a) ( x − 2 )^2 − ( x − 3 ) ( x + 3 ) = 6. b) 2x ( x − 3 ) − 5 ( 3 − x ) = 0.
Hướng dẫn giải
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 \(x \in \left\{ {3; - \frac{5}{2}} \right\}.\) |
c) \(2{x^2} - x - 6 = 0\)
\(2{x^2} - 4x + 3x - 6 = 0\)
\(2x\left( {x - 2} \right) + 3\left( {x - 2} \right) = 0\)
\(\left( {x - 2} \right)\left( {2x + 3} \right) = 0\)
Suy ra \(x - 2 = 0\) hoặc \(2x + 3 = 0\)
\(x = 2\) hoặc \(2x = - 3\)
\(x = 2\) hoặc \(x = - \frac{3}{2}.\)
Vậy \(x \in \left\{ {2; - \frac{3}{2}} \right\}.\)