Question

Tìm x để  \(\frac{3\left(x-2\right)\left(x+2\right)}{x+5}>0\)

Answer 1

Ta có : \(\frac{3\left(x-2\right)\left(x+2\right)}{x+5}=\frac{3\left(x^2-4\right)}{x+5}>0\)

Để \(\frac{3\left(x^2-4\right)}{x+5}>0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3\left(x^2-4\right)< 0;x+5< 0\\3\left(x^2-4\right)>0;x+5>0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x^2< 4;x< -5\\x^2>4;x^2>-5\end{array}\right.\)

TH1 : \(x^2< 4\) , mà \(x< -5\)

 \(\Rightarrow x^2\ge\left(-5\right)^2=25>4\) ( vô lí )

TH2 : \(x^2>4\) ; \(x>-5\)

\(\Rightarrow\left|x\right|>2;x>-5\)

\(\Rightarrow x\in\left\{-4;-3;3;4;5;6\right\}\)

Vậy .............