Tìm x, biết: a) x(3 - 2x) + 2(x^2) = 12. b) x(x - 3) - x + 3 = 0
Hướng dẫn giải
a) \(x\left( {3 - 2x} \right) + 2{x^2} = 12\) \(3x - 2{x^2} + 2{x^2} = 12\) \(3x = 12\) \(x = 4\) Vậy \(x = 4\). | b) \(x\left( {x - 3} \right) - x + 3 = 0\) \(x\left( {x - 3} \right) - \left( {x - 3} \right) = 0\) \(\left( {x - 3} \right)\left( {x - 1} \right) = 0\) \(x - 3 = 0\) hoặc \(x - 1 = 0\) \(x = 3\) hoặc \(x = 1\) Vậy \(x \in \left\{ {3;\,\,1} \right\}.\) |
c) \({x^3} - 3{x^2} + 3x - 1 = 0\) \({\left( {x - 1} \right)^3} = 0\) \(x - 1 = 0\) \(x = 1\) Vậy \(x = 1\).
| d) \({\left( {3x - 1} \right)^2} - \left( {{x^2} - 2x + 1} \right) = 0\) \({\left( {3x - 1} \right)^2} - {\left( {x - 1} \right)^2} = 0\) \(\left( {3x - 1 - x + 1} \right)\left( {3x - 1 + x - 1} \right) = 0\) \(2x\left( {4x - 2} \right) = 0\) \(2x = 0\) hoặc \(4x - 2 = 0\) \(x = 0\) hoặc \(x = \frac{1}{2}\) Vậy \(x \in \left\{ {0;\,\,\frac{1}{2}} \right\}.\) |