Tìm x , biết: a) 7x^2 + 14x = 0 ; b) x^3 − 7x^2 + 14x − 8 = 0.

a) \(7{x^2} + 14x = 0\)

\(7x\left( {x + 2} \right) = 0\)

Suy ra \(7x = 0\) hoặc \(x + 2 = 0\)

 \(x = 0\) hoặc \(x =  - 2\)

b) \({x^3} - 7{x^2} + 14x - 8 = 0\)

\(\left( {{x^3} - 8} \right) + \left( { - 7{x^2} + 14x} \right) = 0\)

\(\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) - 7x\left( {x - 2} \right) = 0\)

\(\left( {x - 2} \right)\left( {{x^2} + 2x + 4 - 7x} \right) = 0\)

\(\left( {x - 2} \right)\left( {{x^2} - 5x + 4} \right) = 0\)

\(\left( {x - 2} \right)\left( {{x^2} - x - 4x + 4} \right) = 0\)

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

\(\left( {x - 2} \right)\left[ {x\left( {x - 1} \right) - 4\left( {x - 1} \right)} \right] = 0\)

\(\left( {x - 2} \right)\left( {x - 1} \right)\left( {x - 4} \right) = 0\)

Suy ra \(x - 2 = 0\) hoặc \(x - 1 = 0\) hoặc \(x - 4 = 0\)

\(x = 2\) hoặc \(x = 1\) hoặc \(x = 4.\)

Vậy \(x \in \left\{ {1;2;4} \right\}.\)