Срочно 2 столбик , прошу

№146
$\frac{c-3}{c} - \frac{c^2-9}{c} * \frac{1}{c-3} = \frac{c-3}{c} - \frac{(c-3)(c+3)}{c} * \frac{1}{c-3}= \frac{c-3}{c} - \frac{c+3}{c}= \frac{c-3-c-3}{c}=$ $- \frac{6}{c}$
№147
$\frac{a-1}{a^2} * \frac{ax-a}{a-1} + \frac{1-x}{2a}= \frac{a-1}{a^2} * \frac{a(x-1)}{a-1} + \frac{1-x}{2a}= \frac{x-1}{a} + \frac{1-x}{2a}= \frac{2x-2+1-x}{2a}=$ $\frac{x-1}{2a}$
№148
$\frac{x^2-xy}{y-1} * \frac{y-1}{ x^{2} } + \frac{y-x}{2x} = \frac{x(x-y)}{y-1} * \frac{y-1}{ x^{2} } + \frac{y-x}{2x} = \frac{x-y}{x}+ \frac{y-x}{2x}= \frac{2x-2y+y-x}{2x}=$ $\frac{x-y}{2x}$
№149
$\frac{x}{a} - \frac{x^2-a^2}{a^2}* \frac{a}{x+a} = \frac{x}{a} - \frac{(x-a)(x+a)}{a^2}* \frac{a}{x+a} = \frac{x}{a} - \frac{x-a}{a}= \frac{x-x+a}{a}= \frac{a}{a} =1$