a) Rút gọn
\(Q=\dfrac{a^3-3a^2+3a-1}{a^2-1}\)
= \(\dfrac{a^3-1-3a^2+3a}{\left(a-1\right)\left(a+1\right)}\)
= \(\dfrac{\left(a-1\right)\left(a^2+a+1\right)-3a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}\)
= \(\dfrac{\left(a-1\right)\left(a^2-2a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
= \(\dfrac{\left(a-1\right)^2}{a+1}\)
b)
Tìm giá trị của Q khi |a|=5
**Với a = 5 ta có:
Q= \(\dfrac{\left(5-1\right)^2}{5+1}=\dfrac{4^2}{6}=\dfrac{16}{6}=\dfrac{8}{3}\)
** Với a= -5 ta có:
Q= \(\dfrac{\left(-5-1\right)^2}{-5+1}=\dfrac{\left(-6\right)^2}{-4}=\dfrac{36}{-4}=-9\)
\(\dfrac{a^3-3a^2+3a-1}{a^2-1}=\dfrac{\left(a^3-1\right)-\left(3a^2-3a\right)}{\left(a+1\right)\left(a-1\right)}\)\(\dfrac{\left(a-1\right)\left(a^2+a+1\right)-3a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}=\dfrac{\left(a-1\right)\left(a^2-2a+1\right)}{\left(a-1\right)\left(a+1\right)}=\dfrac{\left(a-1\right)\left(a-1\right)^2}{\left(a-1\right)\left(a+1\right)}\)\(\dfrac{\left(a-1\right)^2}{a+1}\)
Khi |a|=5
a=5\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a-1}=\dfrac{\left(5-1\right)^2}{5-1}=4\)
a=-5\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a-1}=\dfrac{\left(-5-1\right)^2}{-5-1}=-6\)
