Question

Tìm x bt

[x+1/2].[2-x]>0

Answer 1

`[x+1/2][2-x]>0`

`<=>[x+1/2][x-2]<0`

`<=>` $ \begin{cases}x+\dfrac12>0\\x-2<0\\\end{cases}$

`<=>` $ \begin{cases}x>-\dfrac12\\x<2\\\end{cases}$

`<=>-1/2<x<2`

Nếu cho x nguyên 

`=>x in {0;1}`

Answer 2

Ta có: \(\left(x+\dfrac{1}{2}\right)\left(2-x\right)>0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)\left(x-2\right)< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}>0\\x-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{1}{2}\\x< 2\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}< x< 2\)