04.22. a) 4a + a-1', 6) a-1-2-3a. a-2'

a) $$4a + \frac{1}{a-1} = \frac{4a \cdot (a-1) + 1}{a-1} = \frac{4a^2 - 4a + 1}{a-1}$$ б) $$a - 1 - \frac{2-3a}{a-2} = \frac{(a-1)(a-2) - (2-3a)}{a-2} = \frac{a^2 - 2a - a + 2 - 2 + 3a}{a-2} = \frac{a^2}{a-2}$$ Ответ: a) $$\frac{4a^2 - 4a + 1}{a-1}$$; б) $$\frac{a^2}{a-2}$$