Tìm x , biết: a) 2x^2 − 8x = 0. b) ( x + 2 )^2 − x ( x − 1 ) = 10.

a) \[2{x^2} - 8x = 0\]

\[2x\left( {x - 4} \right) = 0\]

Suy ra \[2x = 0\] hoặc \[x - 4 = 0\]

 \[x = 0\] hoặc \[x = 4\]

Vậy \(x \in \left\{ {0;\,\,4} \right\}\).

b) \({\left( {x + 2} \right)^2} - x\left( {x - 1} \right) = 10\)

\({x^2} + 4x + 4 - {x^2} + x = 10\)

\[\left( {{x^2} - {x^2}} \right) + \left( {4x + x} \right) = 10 - 4\]

\(5x = 6\)

\(x = \frac{6}{5}\)

Vậy\(x = \frac{6}{5}\).

c) \[{x^3} - 6{x^2} + 9x = 0\]

\[x\left( {{x^2} - 6x + 9} \right) = 0\]

\[x{\left( {x - 3} \right)^2} = 0\]

Suy ra \[x = 0\] hoặc \[{\left( {x - 3} \right)^2} = 0{\rm{ }}\]

 \[x = 0\] hoặc \[x - 3 = 0{\rm{ }}\]

 \[x = 0\] hoặc \[x = 3\]

Vậy \(x \in \left\{ {0;\,\,3} \right\}\).