Đáp án:
`a,S= (-\infty;-3]∪[1/2;+\infty)`
`b, S=(3/2;4)`
`c, S=[-2;-1]`
Giải thích các bước giải:
`a, (x+3)(2x-1)≥0`
Xét `f(x) = (x+3)(2x-1)`
`x+3=0=> x =-3`
`2x -1=0=> x =1/2`
Bảng xét dấu:
$\begin{array} {|c|cc|} \hline x&-\infty&&-3&&1/2&&+\infty\\\hline x+3&&-&0&+&|&+&&\\\hline 2x-1&&-&|&-&0&+&&\\\hline f(x)&&+&0&-&0&+\\\hline\end{array}$
`=> f(x)≥0 <=> x \in (-\infty;-3]∪[1/2;+\infty)`
Vậy `S= (-\infty;-3]∪[1/2;+\infty)`
____________
`b, (-2x +3)(4-x)<0`
Xét `f(x) = (-2x +3)(4-x)`
`-2x+3=0=> x=3/2`
`4-x=0=> x=4`
Bảng xét dấu:
$\begin{array} {|c|cc|} \hline x&-\infty&&3/2&&4&&+\infty\\\hline -2x+3&&+&0&-&|&-&&\\\hline 4-x&&+&|&+&0&-&&\\\hline f(x)&&+&0&-&0&+\\\hline\end{array}$
`=> f(x)<0<=> x \in (3/2;4)`
Vậy `S=(3/2;4)`
__________
`c, (x+2)/(x+1)≤0`
Xét `f(x)=(x+2)/(x+1)`
`x+2=0=> x=-2`
`x+1=0=> x=-1`
Bảng xét dấu: $\begin{array} {|c|cc|} \hline x&-\infty&&-2&&-1&&+\infty\\\hline x+2&&-&0&+&|&+&&\\\hline x+1&&-&|&-&0&+&&\\\hline f(x)&&+&0&-&0&+\\\hline\end{array}$
`=> f(x) ≤0 <=> x\in [-2;-1]`
Vậy `S=[-2;-1]`
