1. $f(x)=(5x+1)(4-x)>0$
Ta có: $5x+1=0 \Leftrightarrow x=-\frac{1}{5}$
$4-x=0 \Leftrightarrow x=4$
$*BXD:$
x -∞ -1/5 4 +∞
f(x) - 0 + 0 -
Dựa vào BXD:
$\Leftrightarrow -\frac{1}{5}<x<4$ thoả ycbt.
2. $f(x)=x^2-4x+3≥0$
Ta có: $x^2-4x+3=0$
$\Leftrightarrow \left[ \begin{array}{l}x=1\\x=3\end{array} \right.$
$*BXD:$
x -∞ 1 3 +∞
f(x). + 0 - 0 +
Dựa vào BXD:
$\Leftrightarrow x≤1$ hay $3≤x$ thoả ycbt.