a) \(\left(x+\frac{1}{9}\right)\left(2x-5\right)< 0\)
TH1 : \(\hept{\begin{cases}x+\frac{1}{9}>0\\2x-5< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>\frac{-1}{9}\\x< \frac{5}{2}\end{cases}}\)
\(\Leftrightarrow\frac{-1}{9}< x< \frac{5}{2}\)( thỏa )
TH2 : \(\hept{\begin{cases}x+\frac{1}{9}< 0\\2x-5>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x< -\frac{1}{9}\\x>\frac{5}{2}\end{cases}}\)
\(\Leftrightarrow\frac{5}{2}< x< -\frac{1}{9}\)( loại )
Vậy....
b) \(x^2-6x+9< 0\)
\(\Leftrightarrow\left(x-3\right)^2< 0\)( vô lý )
Vậy bpt vô nghiệm
a \(\left(x+\frac{1}{9}\right).\left(2x-5\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{9}>0\\2x-5< 0\end{cases}\text{hoặc}\hept{\begin{cases}x+\frac{1}{9}< 0\\2x-5>0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x>-\frac{1}{9}\\x< \frac{5}{2}\end{cases}\text{hoặc}\hept{\begin{cases}x< -\frac{1}{9}\\x>\frac{5}{2}\end{cases}}\left(loai\right)}\)
\(\Rightarrow-\frac{1}{9}< x< \frac{5}{2}\)
b \(x^2-6x+9< 0\Rightarrow x^2-3x-3x+9< 0\)
\(\Rightarrow x.\left(x-3\right)-3.\left(x-3\right)< 0\Rightarrow\left(x-3\right)^2< 0\left(\text{loại vì }\left(x-3\right)^2\ge0\right)\)
