Tìm x biết: x^3 + 9x^2 + 27x + 19 = 0

a) \[{x^3} + 9{x^2} + 27x + 19 = 0\]

\[{x^3} + 9{x^2} + 27x + 27 - 8 = 0\]

\[{\left( {x + 3} \right)^3} = 8\]

Suy ra \[x + 3 = 2\]

\(x =  - 1.\)

Vậy \(x =  - 1.\)

b) \[25{\left( {x + 3} \right)^2} + \left( {1-5x} \right)\left( {1 + 5x} \right) = 8\]

\(25\left( {{x^2} + 6x + 9} \right) + \left[ {{1^2} - {{\left( {5x} \right)}^2}} \right] = 8\)

\[25{x^2} + 150x + 225 + 1 - 25{x^2} = 8\]

\[150x =  - 218\]

\(x =  - \frac{{109}}{{75}}.\)

Vậy \(x =  - \frac{{109}}{{75}}.\)