Question

BT:Tìm x,biết:

a) (x²+1).(2x+1).(2x-1)>0

b) (x+1).(x-3)>0

Answer 1

a) \(\left(x^2+1\right)\left(2x+1\right)\left(2x-1\right)>0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)>0\)

\(\Leftrightarrow\begin{cases}2x-1>0\\2x+1>0\end{cases}\) hoặc \(\begin{cases}2x-1< 0\\2x+1< 0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-\frac{1}{2}\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -\frac{1}{2}\end{cases}\)

\(\Leftrightarrow x>\frac{1}{2}\) hoặc \(x< -\frac{1}{2}\)

b) \(\left(x+1\right)\left(x-3\right)>0\)

\(\Leftrightarrow\begin{cases}x+1>0\\x-3>0\end{cases}\) hoặc \(\begin{cases}x+1< 0\\x-3< 0\end{cases}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x>3\\x< -1\end{array}\right.\)