Tìm x , biết: a) 9x ( x − 1 ) − ( 3 x − 1 )^2 = 0. b) x^3 − 4x = 0. c) x^2 + 5x = 6.

a) \(9x\left( {x - 1} \right) - {\left( {3x - 1} \right)^2} = 0\)

\(9{x^2} - 9x - \left( {9{x^2} - 6x + 1} \right) = 0\)

\(9{x^2} - 9x - 9{x^2} + 6x - 1 = 0\)

\(\left( {9{x^2} - 9{x^2}} \right) + \left( { - 9x + 6x} \right) - 1 = 0\)

\( - 3x = 1\)

\(x =  - \frac{1}{3}.\)

Vậy \(x =  - \frac{1}{3}.\)

b) \({x^3} - 4x = 0\)

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

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

Suy ra \(x = 0\) hoặc \(x - 2 = 0\) hoặc \(x + 2 = 0\)

 \(x = 0\) hoặc \(x = 2\) hoặc \[x =  - 2.\]

Vậy \(x \in \left\{ {0;2; - 2} \right\}.\)

c) \({x^2} + 5x = 6\)

\({x^2} + 5x - 6 = 0\)

\({x^2} - x + 6x - 6 = 0\)

\(x\left( {x - 1} \right) + 6\left( {x - 1} \right) = 0\)

\(\left( {x - 1} \right)\left( {x + 6} \right) = 0\)

Suy ra \(x - 1 = 0\) hoặc \(x + 6 = 0\)

\(x = 1\) hoặc \(x =  - 6.\)

Vậy \(x \in \left\{ {1; - 6} \right\}.\)