Answer:
Step-by-step explanation:
[tex]\frac{(2x+1)^2}{(2x+1)(x-3)} -\frac{5}{x^2-5x+6} \\=\frac{2x+1}{x-3} -\frac{5}{x^2-2x-3x+6} \\=\frac{2x+1}{x-3} -\frac{5}{x(x-2)-3(x-2) } \\=\frac{2x+1}{x-3} -\frac{5}{(x-2)(x-3)}\\=\frac{(2x+1)(x-2)-5}{(x-2)(x-3)}\\=\frac{2x^2-3x-2-5}{x^2-5x+6} \\\[=\frac{2x^2-3x-7}{x^2-5x+6}[/tex]
