Question

Tìm x biết: \(\left(x-3\right)\left(x+\dfrac{1}{2}\right)>0\)

Answer 1

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

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\Rightarrow x>3\\x+\dfrac{1}{2}< 0\Rightarrow x< -\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\Rightarrow x< 3\\x+\dfrac{1}{2}>0\Rightarrow x>-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-\dfrac{1}{2}< x< 3\)

Answer 2

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

TH1 :

\(\Leftrightarrow\left[{}\begin{matrix}x-3< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x< 3\\x< \dfrac{-1}{2}\end{matrix}\right.\) \(\Leftrightarrow x< \dfrac{-1}{2}\)

Th2 :

\(\Leftrightarrow\left[{}\begin{matrix}x-3>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x>\dfrac{-1}{2}\end{matrix}\right.\) \(\Leftrightarrow x>3\)

Vậy ...

Answer 3

Ta có : $x+\dfrac{1}{2}>x-3$

* Xét $x+\dfrac{1}{2};x-3>0$

\(=>\left\{{}\begin{matrix}x-3>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\)

\(=>\left\{{}\begin{matrix}x>3\\x>\dfrac{-1}{2}\end{matrix}\right.\)

$=>x>3$

* Xét $x-3;x+\dfrac{1}{2}<0$

\(=>\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\)

\(=>\left\{{}\begin{matrix}x< 3\\x< \dfrac{-1}{2}\end{matrix}\right.\)

$=>x<\dfrac{-1}{2}$

Vậy để $(x-3)(x+\dfrac{1}{2}>0$ thì \(\left[{}\begin{matrix}x>3\\x< \dfrac{-1}{2}\end{matrix}\right.\).

Answer 4

Mk sửa lại nhé:

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

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\Rightarrow x>3\\x+\dfrac{1}{2}>0\Rightarrow x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\Rightarrow x< 3\\x+\dfrac{1}{2}< 0\Rightarrow x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy xảy ra khi:

\(x>-\dfrac{1}{2};x< 3\)