2.176. $\frac{a^{2}-4-|a-2|}{a^{3}+2 a^{2}-5 a-6}$.

2.176. $\frac{a^{2}-4-|a-2|}{a^{3}+2 a^{2}-5 a-6}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}a \neq 2, \\ a \neq-3, \\ a \neq-1 .\end{array}\right.$

$$
\begin{aligned}
& \frac{a^{2}-4-|a-2|}{a^{3}+2 a^{2}-5 a-6}=\frac{(a-2)(a+2)-|a-2|}{(a-2)(a+3)(a+1)}= \\
& =\left\{\begin{aligned}
& \frac{(a-2)(a+2)+(a-2)}{(a-2)(a+3)(a+1)}=\frac{(a-2)(a+2+1)}{(a-2)(a+3)(a+1)}=\frac{a+3}{(a+3)(a+1)}=\frac{1}{a+1} \\
& \frac{(a-2)(a+2)-(a-2)}{(a-2)(a+3)(a+1)}=\frac{(a-2)(a+2-1)}{(a-2)(a+3)(a+1)}= \frac{a+1}{(a+3)(a+1)}=\frac{1}{a+3} \\
& \text { for } a \in(2 ; \infty) .
\end{aligned}\right.
\end{aligned}
$$

Answer: $\frac{1}{a+1}$ for $a \in(-\infty ;-3) \cup(-3 ;-1) \cup(-1 ; 2) ; \frac{1}{a+3} \quad$ for $a \in(2 ; \infty)$.

$$
\text { 2.177. } \frac{2 x-x \cdot|x-1|+x \cdot|x|+3}{|x|+x^{2}}
$$

## Solution.

$$
\begin{aligned}
& \text { Domain of definition: }\left\{\begin{array}{l}
x \neq 0, \\
x \neq 1, \\
x \neq-1
\end{array}\right. \\
& \frac{2 x-x \cdot|x-1|+x \cdot|x|+3}{|x|+x^{2}}= \\
& =\left\{\begin{array}{l}
\frac{2 x+x(x-1)-x^{2}+3}{-x+x^{2}}=\frac{x+3}{x^{2}-x} \text { for } x \in(-\infty ; 0), \\
\frac{2 x+x(x-1)+x^{2}+3}{x^{2}+x}=\frac{2 x^{2}+x+3}{x^{2}+x} \text { for } x \in(0 ; 1) \\
\frac{2 x-x(x-1)+x^{2}+3}{x+x^{2}}=\frac{3(x+1)}{x(x+1)}=\frac{3}{x} \text { for } x \in[1 ; \infty)
\end{array}\right.
\end{aligned}
$$

Answer: $\frac{x+3}{x^{2}-x}$ for $x \in(-\infty ; 0) ; \frac{2 x^{2}+x+3}{x^{2}+x}$ for $x \in(0 ; 1) ; \frac{3}{x}$ for $x \in[1 ; \infty)$.