\(a,Q=\frac{a^3-3a^2+3a-1}{a^2-1}=\frac{\left(a-1\right)^3}{\left(a-1\right)\left(a+1\right)}=\frac{\left(a-1\right)^2}{a+1}.\)
b, ta có : \(/a/=5\Rightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)
thay a = -5 vào Q
\(\Rightarrow Q=\frac{\left(-5-1\right)^2}{-5+1}=\frac{36}{-4}=-9\)
thay a = 5 vào Q
\(\Rightarrow Q=\frac{\left(5-1\right)^2}{5+1}=\frac{16}{6}=\frac{8}{3}\)
KL : Q = 8/3 tại x=5
\(\text{Đ}K\text{X}\text{Đ}:a\ne1\)
a) Ta có: \(Q=\frac{a^3-3a^2+3a-1}{a^2-1}=\frac{\left(a-1\right)^3}{\left(a-1\right)\left(a+1\right)}\)
Vậy ....
b) Ta có: \(\left|a\right|=5\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)
Với a=5 ta có: \(Q=\frac{\left(5-1\right)^2}{5+1}=\frac{16}{6}=\frac{8}{3}\)
Với a=-5 ta có: \(Q=\frac{\left(-5-1\right)^2}{-5+1}=\frac{36}{-4}=-9\)