\( x^{2}-3 x+\frac{4}{x-2}=\frac{4}{x-2}-2 \) Ombem: \( \left\lvert\, \begin{array}{l}\left|x^{2}+5 x-3\right|=3\end{array}\right. \)

To solve the equation \( x^{2}-3 x+\frac{4}{x-2}=\frac{4}{x-2}-2 \), first, we notice that \(\frac{4}{x-2}\) is common on both sides. By simplifying, we get \( x^{2} - 3x = -2 \). Rearranging gives \( x^{2} - 3x + 2 = 0\), which factors to \((x-1)(x-2)=0\). Therefore, the solutions are \( x = 1 \) and \( x = 2 \). However, \( x = 2 \) is extraneous due to the denominator, leaving \( x = 1 \) as the only solution. For the absolute value equation \(\left|\, \begin{array}{l}\left|x^{2}+5 x-3\right|=3\end{array}\right.\), we can split it into two cases: \( x^{2}+5x-3=3 \) and \( x^{2}+5x-3=-3 \). Solving these gives us \( x^{2}+5x-6=0 \) (factoring to \( (x-1)(x+6)=0 \)) and \( x^{2}+5x=0 \) (factoring gives \( x(x+5)=0 \)). The solutions are \( x=1, -6, 0, -5 \). So the complete solution set includes \( x = 1, -6, 0, -5\).