Tìm x biết : a) x ( 3-2x ) + 2x^2 =12

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 \(x \in \left\{ {0;\,\,\frac{1}{2}} \right\}.\)