Thực hiện phép tính:

j) \(1 + \frac{{{x^3} - x}}{{{x^2} + 1}} \cdot \left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right).\)

b) \(\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) + x\left( {3 - {x^2}} \right) = x\)

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

\(2x - 8 = 0\)

\(2x = 8\)

\(x = 4\).

g) \[4{\left( {x - 2} \right)^2} = 25{\left( {1 - 3x} \right)^2}\]

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

\[{\left( {2x - 4} \right)^2} - {\left( {5 - 15x} \right)^2} = 0\]

\[\left( {2x - 4 + 5 - 15x} \right)\left( {2x - 4 - 5 + 15x} \right) = 0\]

\[\left( { - 13x + 1} \right)\left( {17x - 9} \right) = 0\]

\[ - 13x + 1 = 0\] hoặc \[17x - 9 = 0\]

\( - 13x = - 1\) hoặc \(17x = 9\)

\(x = \frac{1}{{13}}\) hoặc \(x = \frac{9}{{17}}\).