Question

\(choQ=\frac{a^3-3a^2+3a-1}{a^2-1}\)

a) rút gọn Q

b) tìm a để Q<0

Answer 1

a ) \(Q=\frac{\left(a^3-1\right)-3a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}=\frac{\left(a-1\right)\left(a^2+a+1\right)-3a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}=\frac{\left(a-1\right)\left(a^2-2a+1\right)}{\left(a-1\right)\left(a+1\right)}\)

\(=\frac{\left(a-1\right)\left(a-1\right)^2}{\left(a-1\right)\left(a+1\right)}=\frac{\left(a-1\right)^2}{a+1}\)

b ) Để \(Q< 0\) \(\Leftrightarrow\frac{\left(a-1\right)^2}{a+1}< 0\)

Mà \(\left(a-1\right)^2\ge0\) nên \(a+1< 0\Rightarrow a< -1\)

Vậy \(a< -1\)

Answer 2

dễ lắm đấy

Answer 3

giải đi