Bài 3:
a) Để $\left ( x-\frac{1}{5} \right )(2x+1)$ có giá trị âm
$\left ( x-\frac{1}{5} \right )(2x+1)<0$
$\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
x-\frac{1}{5}<0\\
2x+1>0
\end{matrix}\right.\\\left\{\begin{matrix}
x-\frac{1}{5}>0\\
2x+1<0
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
x<\frac{1}{5}\\
x>-\frac{1}{2}
\end{matrix}\right.\\\left\{\begin{matrix}
x>\frac{1}{5}\\
x<-\frac{1}{2}
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x\in \left \{ -\frac{1}{2};\frac{1}{5} \right \}\\x\in \varnothing\end{array} \right.$
Vậy $x\in \left \{ -\frac{1}{2};\frac{1}{5} \right \}$
b) Để $\left ( \frac{1}{2}x-1 \right )\left ( \frac{1}{3}x-2 \right )$ có giá trị âm
$\Leftrightarrow \left ( \frac{1}{2}x-1 \right )\left ( \frac{1}{3}x-2 \right )<0$
$\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
\frac{1}{2}x-1<0\\
\frac{1}{3}x-2>0
\end{matrix}\right.\\\left\{\begin{matrix}
\frac{1}{2}x-1>0\\
\frac{1}{3}x-2<0
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
x<2\\
x>6
\end{matrix}\right.\\\left\{\begin{matrix}
x>2\\
x<6
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x\in \varnothing\\x\in \left \{ 2,6 \right \}\end{array} \right.$
Vậy $x\in \left \{ 2,6 \right \}$
c) Để $\frac{x-2}{x-6}$ có giá trị âm
$\Leftrightarrow \frac{x-2}{x-6}<0$
$\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
x-2<0\\
x-6>0
\end{matrix}\right.\\\left\{\begin{matrix}
x-2>0\\
x-6<0
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
x<2\\
x>6
\end{matrix}\right.\\\left\{\begin{matrix}
x>2\\
x<6
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x\in \varnothing\\x\in \left \{ 2;6 \right \}\end{array} \right.$
Vậy $x\in \left \{ 2;6 \right \}$
d) Để $x^{2}-\frac{2}{5}x$ có giá trị âm
$\Leftrightarrow x^{2}-\frac{2}{5}x<0$
$\Leftrightarrow 5x^{2}-2x<0\\\Leftrightarrow x(5x-2)<0\\\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
x<0\\
5x-2>0
\end{matrix}\right.\\\left\{\begin{matrix}
x>0\\
5x-2<0
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
x<0\\
x>\frac{2}{5}
\end{matrix}\right.\\\left\{\begin{matrix}
x>0\\
x<\frac{2}{5}
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x\in \varnothing\\x\in \left \{ 0;\frac{2}{5} \right \}\end{array} \right.$
Vậy $x\in \left \{ 0;\frac{2}{5} \right \}$
