Question

[Nguyen Thi Minh Tam 1]

Tìm x biết :(x-3)(x+1/2)>0

Answer 1

[Yêu nè]

Linh tinh thôi đó

\(\left(x-3\right)\left(x+\frac{1}{2}\right)>0\)

\(\Leftrightarrow\hept{\begin{cases}x-3>0\\x+\frac{1}{2}>0\end{cases}}\)  hoặc \(\hept{\begin{cases}x-3< 0\\x+\frac{1}{2}< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>3\\x>\frac{-1}{2}\end{cases}}\)  hoặc \(\hept{\begin{cases}x< 3\\x< \frac{-1}{2}\end{cases}}\)

\(\Leftrightarrow x>3\) hoặc \(x< \frac{-1}{2}\)

\(\Leftrightarrow-\frac{1}{2}< x< 3\)

Vậy \(-\frac{1}{2}< x< 3\)

Answer 2

[Trần Linh]

\(\left(x-3\right)\left(x+\frac{1}{2}\right)>0\)

\(\Leftrightarrow\)\(\left(x-3\right)\)và \(\left(x+\frac{1}{2}\right)\)cùng dấu

TH1: \(x-3>0\)và \(x+\frac{1}{2}>0\)

\(\Leftrightarrow\)\(x>3\).    .  và \(x>\frac{-1}{2}\)

\(\Leftrightarrow x>3\)

TH2: \(x-3< 0\)và \(x+\frac{1}{2}< 0\)

\(\Leftrightarrow x< 3\)           và \(x< \frac{-1}{2}\)

\(\Leftrightarrow x< \frac{-1}{2}\)

Vậy \(x>3\)và \(x< \frac{-1}{2}\)thì\(\left(x-3\right)\left(x+\frac{1}{2}\right)>0\)