Tìm \[x,\] biết

a) \({\left( {x - 2} \right)^2} - x \cdot \left( {x - 1} \right) = 9.\)         

b) \[\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - {x^3} + 2x =  - 35.\]          

c) \[{x^2} - 4x = 5.\]

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

\(\,\,\,\,\,\,\,\,\,{x^2} - 4x + 4 - {x^2} + x = 9\)

\(\,\,\,\,\,\,\,\,\, - 3x + 4 = 9\)

\(\,\,\,\,\,\,\,\,\, - 3x = 5\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{ - 5}}{3}\)

Vậy \(x = \frac{{ - 5}}{3}.\)

b) \[\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - {x^3} + 2x = - 35\]

\[{x^3} + {1^3} - {x^3} + 2x = - 35\]

\[2x + 1 = - 35\]

\[2x = - 36\]

 \[x = - 18\]

Vậy \[x = - 18.\]

c) \[{x^2} - 4x = 5\]

\[{x^2} - 4x + 4 = 5 + 4\]

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

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

Trường hợp 1: \[x - 2 = 3\]

\[x = 5\]

Trường hợp 2: \[x - 2 = - 3\]

\[x = - 1\]

Vậy \(x \in \left\{ {5; - 1} \right\}.\)