Tìm x , biết: a) ( 2x − 1 )^2 − 25 = 0. b) x^3 + 27 + ( x + 3 ) ( x − 9 ) = 0.

a) \({\left( {2x - 1} \right)^2} - 25 = 0\)

\({\left( {2x - 1} \right)^2} - {5^2} = 0\)

\(\left( {2x - 1 - 5} \right)\left( {2x - 1 + 5} \right) = 0\)

\(\left( {2x - 6} \right)\left( {2x + 4} \right) = 0\)

Trường hợp 1:

\(2x - 6 = 0\)

\(2x = 6\)

\(x = 3\)

Trường hợp 2:

\(2x + 4 = 0\)

\(2x = - 4\)

\(x = - 2\)

Vậy \(x \in \left\{ {3; - 2} \right\}.\)

b) \({x^3} + 27 + \left( {x + 3} \right)\left( {x - 9} \right) = 0\)

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

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

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

\(\left( {x + 3} \right)x\left( {x - 2} \right) = 0\)

Trường hợp 1:

\(x + 3 = 0\)

\(x = - 3\)

Trường hợp 2:

\(x = 0\)

Trường hợp 3:

\(x - 2 = 0\)

\(x = 2\)

Vậy \(x \in \left\{ { - 3;0;2} \right\}.\)