Tìm x, biết: 2(3x - 5) + 2(1 - x)=  - 4

a) \(2\left( {3x - 5} \right) + 2\left( {1 - x} \right) = - 4\)

\(6x - 10 + 2 - 2x = - 4\)

\(6x - 2x = 10 - 2 - 4\)

\(4x = 4\)

\(x = 1\)

Vậy \(x = 1\).

b) \(6x\left( {x + 1} \right) - \left( {2x - 5} \right)\left( {3x + 2} \right) = - 5\)

\[\left( {6{x^2} + 6x} \right) - \left( {6{x^2} + 4x - 15x - 10} \right) = - 5\]

\[6{x^2} + 6x - 6{x^2} + 11x + 10 = - 5\]

\[6{x^2} - 6{x^2} + 6x + 11x = - 10 - 5\]

\[17x = - 15\].

\[x = - \frac{{15}}{{17}}\].

Vậy \[x = - \frac{{15}}{{17}}\].