Question

Tìm số nguyên x để:

\(\left(2x+1\right)\)\(.\)\(\left(x-\frac{1}{2}\right)\)\(< 0\)

Answer 1

\(\left(2x+1\right).\left(x-\frac{1}{2}\right)< 0\)

TH1 :

\(\hept{\begin{cases}2x+1< 0\\x-\frac{1}{2}>0\end{cases}\Rightarrow\hept{\begin{cases}x< \frac{-1}{2}\\x>\frac{1}{2}\end{cases}\Rightarrow}x\in\theta}\)

TH2:

\(\hept{\begin{cases}2x+1>0\\x-\frac{1}{2}< 0\end{cases}\Rightarrow\hept{\begin{cases}x>\frac{-1}{2}\\x< \frac{1}{2}\end{cases}\Rightarrow}\frac{-1}{2}< x< \frac{1}{2}}\)

Vậy \(\frac{-1}{2}< x< \frac{1}{2}\)