Помогите пожалуйста!

Решите задачу:

$\displaystyle\tt 1) \ \ \ \ \frac{2x^2+5x-3}{x-1}= \frac{2x^2-2x+2x+5x-3-4+4}{x-1}=\\\\\\{} \ \ \ =\frac{2x^2-2x}{x-1}+\frac{7x-7}{x-1}+\frac{4}{x-1}=\frac{2x(x-1)}{x-1}+\frac{7(x-1)}{x-1}+\frac{4}{x-1}=\\\\\\{} \ \ \ =2x+7+\frac{4}{x-1}\\\\\\\\ 2) \ \ \ \ \frac{4x^2-2x+3}{x-1}=\frac{4x^2-2x-2x+2x+3-5+5}{x-1}=\\\\\\{} \ \ \ =\frac{4x^2-4x}{x-1}+\frac{2x-2}{x-1}+\frac{5}{x-1}=\frac{4x(x-1)}{x-1}+\frac{2(x-1)}{x-1}+\frac{5}{x-1}=\\\\\\{} \ \ \ =4x+2+\frac{5}{x-1}$