Để : \(\left(x+1\right).\left(x-1\right)< 0\)
Thì 1 trong hai số phải < 0
Xảy ra hai trường hợp:
\(\left(1\right)\begin{cases}x+1< 0\\x-1>0\end{cases}\Rightarrow\begin{cases}x< -1\\x>1\end{cases}\Rightarrow-1< x< 1\)
\(\left(2\right)\begin{cases}x+1>0\\x-1< 0\end{cases}\Rightarrow\begin{cases}x>-1\\x< 1\end{cases}\Rightarrow x\in O\)
Để : \(\left(x+1\right).\left(x-1\right)=0\)
\(\Rightarrow\begin{cases}x+1=0\\x-1=0\end{cases}\Rightarrow\begin{cases}x=-1\\x=1\end{cases}\)
\(\left(x+1\right)\left(x-1\right)\le0\)
\(\Leftrightarrow x^2-1\le0\)
\(\Leftrightarrow x^2\le1\)
\(\Leftrightarrow\left|x\right|\le1\)
\(\Leftrightarrow-1\le x\le1\)
Ta có
\(\left(x+1\right)\left(x-1\right)=x^2-1\)
\(\Rightarrow x^2-1\le0\)
\(\Rightarrow x^2\le1\)
Mà \(x^2\ge0\) với mọi x
\(\Rightarrow\left[\begin{array}{nghiempt}x^2=0\\x^2=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\\left[\begin{array}{nghiempt}x=1\\x=-1\end{array}\right.\end{array}\right.\)
Vậy x=0 ; x=1 ; x=-1
