Because $\displaystyle{{x}}^{{2}}-{5}{x}+{6}={\left({x}-{2}\right)}{\left({x}-{3}\right)}\usepackage{cmbright}\Huge\$ , this is the least common denominator.

$\displaystyle\frac{{{3}{x}-{2}}}{{{x}-{2}}}-\frac{{{x}-{1}}}{{{{x}}^{{2}}-{5}{x}+{6}}}=\frac{{{\left({3}{x}-{2}\right)}{\left({x}-{3}\right)}}}{{{\left({x}-{2}\right)}{\left({x}-{3}\right)}}}-\frac{{{x}-{1}}}{{{\left({x}-{2}\right)}{\left({x}-{3}\right)}}}=\frac{{{3}{{x}}^{{2}}-{11}{x}+{6}}}{{{{x}}^{{2}}-{5}{x}+{6}}}-\frac{{{\left({x}-{1}\right)}}}{{{{x}}^{{2}}-{5}{x}+{6}}}\usepackage{cmbright}\Huge\$

$\displaystyle=\frac{{{3}{{x}}^{{2}}-{11}{x}+{6}-{x}+{1}}}{{{{x}}^{{2}}-{5}{x}+{6}}}=\frac{{{3}{{x}}^{{2}}-{12}{x}+{7}}}{{{{x}}^{{2}}-{5}{x}+{6}}}\usepackage{cmbright}\Huge\$