Đáp án:
`a)S={x|x\inQQ;1/12<x<5}`
`b)S={x|x\inQQ;-4<x<1/2}`
Giải thích các bước giải:
`a) (5-x)(3x-1/4)>0`
`<=>`\(\left[ \begin{array}{l}\begin{cases}5-x>0\\3x-\dfrac{1}{4}>0\end{cases}\\\begin{cases}5-x<0\\3x-\dfrac{1}{4}<0\end{cases}\end{array} \right.\)`<=>`\(\left[ \begin{array}{l}\begin{cases}x<5\\x>\dfrac{1}{12}\end{cases}\\\begin{cases}x>5\\x<\dfrac{1}{12}\end{cases}\end{array}\right.\) `<=>1/12<x<5`
Vậy `S={x|x\inQQ;1/12<x<5}`
`b)` $\dfrac{x-\dfrac{1}{2}}{4+x}<0$ ĐK:`x\ne-4`
`<=>`\(\left[ \begin{array}{l}\begin{cases}x-\dfrac{1}{2}>0\\x+4<0\end{cases}\\\begin{cases}x-\dfrac{1}{2}<0\\x+4>0\end{cases}\end{array} \right.\)`<=>`\(\left[ \begin{array}{l}\begin{cases}x>\dfrac{1}{2}\\x<-4\end{cases}\\\begin{cases}x<\dfrac{1}{2}\\x> -4\end{cases}\end{array}\right.\) `<=>-4<x<1/2`
Vậy `S={x|x\inQQ;-4<x<1/2}`
