Thực hiện phép tính:

h) \[\frac{1}{{x - y}} + \frac{{3xy}}{{{y^3} - {x^3}}} + \frac{{x - y}}{{{x^2} + xy + {y^2}}}.\]

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

\( = 1 + \frac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}} \cdot \left[ {\frac{1}{{1 - x}} - \frac{1}{{\left( {1 - x} \right)\left( {1 + x} \right)}}} \right]\)

\( = 1 - \frac{{x\left( {1 - {x^2}} \right)}}{{{x^2} + 1}} \cdot \frac{{1 + x - 1}}{{\left( {1 - x} \right)\left( {1 + x} \right)}}\)
\( = 1 - \frac{{x\left( {1 - x} \right)\left( {1 + x} \right)}}{{{x^2} + 1}} \cdot \frac{x}{{\left( {1 - x} \right)\left( {1 + x} \right)}}\)

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

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

\(2x - 8 = 0\)

\(2x = 8\)

\(x = 4\).