a) \(2\left( {x + 1} \right) = \left( {5x - 1} \right)\left( {x + 1} \right);\)
\(\begin{array}{l}2\left( {x + 1} \right) = \left( {5x - 1} \right)\left( {x + 1} \right)\\2\left( {x + 1} \right) - \left( {5x - 1} \right)\left( {x + 1} \right) = 0\\\left( {x + 1} \right)\left( {2 - 5x + 1} \right) = 0\\\left( {x + 1} \right)\left( {3 - 5x} \right) = 0\\TH1:x + 1 = 0\\x = - 1\\TH2:3 - 5x = 0\\x = \frac{3}{5}\end{array}\)
Vậy \(x \in \left\{ { - 1;\frac{3}{5}} \right\}.\)
b) \(\left( { - 4x + 3x} \right)x = \left( {2x + 5} \right)x.\)
\(\begin{array}{l}\left( { - 4x + 3x} \right)x - \left( {2x + 5} \right)x = 0\\x\left( { - 4x + 3x - 2x - 5} \right) = 0\\x\left( { - 3x - 5} \right) = 0\\TH1:x = 0\\TH2:x = \frac{{ - 5}}{3}\end{array}\)
Vậy \(x \in \left\{ {0;\frac{{ - 5}}{3}} \right\}.\)