\(Đkxđ:a\ne1\)
\(a,Q=\frac{a^3-3a^2+3a-1}{a^2-1}=\frac{a^3-1-3a^2+3a}{a^2-1}=\frac{\left(a-1\right)\left(a^2+a+1\right)-3a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}=\frac{\left(a-1\right)\left(a^2-2a+1\right)}{\left(a-1\right)\left(a+1\right)}=\frac{a^2-2a+1}{a+1}\)
\(b,\left|a\right|=5\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)
Khi \(=5\) thì: \(Q=\frac{5^2-5.2+1}{5+1}=\frac{8}{3}\)
Khi \(=-5\) thì: \(Q=\frac{\left(-5\right)^2+5.2+1}{-5+1}=-9\)
Lời giải:
a) ĐKXĐ: $a\neq \pm 1$
$Q=\frac{a^3-3a^2+3a-1}{a^2-1}=\frac{(a-1)^3}{(a-1)(a+1)}=\frac{(a-1)^2}{a+1}$
b)
Khi $|a|=5\Rightarrow a=\pm 5$
Nếu $a=5\Rightarrow Q=\frac{(5-1)^2}{5+1}=\frac{8}{3}$
Nếu $a=-5\Rightarrow Q=\frac{(-5-1)^2}{-5+1}=-9$
