Bepaal x en kies die regte antwoord \( 3{x^2} + x - 1 + \frac{1}{{3{x^2} + x - 3}} = 0 \)

\( \begin{align} 3{x^2} + x - 1 + \frac{1}{{3{x^2} + x - 3}} &= 0\\ {\rm{Stel }}\,\,k &= 3{x^2} + x\\ \therefore k - 1 + \frac{1}{{k - 3}} &= 0\\ KGV &= k - 3;BPK:k \ne 3\\ \left( {k - 1} \right)\left( {k - 3} \right) + 1 &= 0\\ {k^2} - 4k + 3 + 1 &= 0\\ {k^2} - 4k + 4 &= 0\\ \left( {k - 2} \right)\left( {k - 2} \right) &= 0\\ \therefore k &= 2\\ {\rm{maar }}\,\,k &= 3{x^2} + x\\ \therefore 3{x^2} + x &= 2\\ 3{x^2} + x - 2 &= 0\\ \left( {3x - 2} \right)\left( {x + 1} \right) &= 0\\ x = \frac{2}{3}\,\,{\rm{of}}\,\,x &= 1 \end{align} \)