mk sẽ làm câu c còn mấy câu còn lại bn lm tương tự cho quen nha :)
c) ta có : \(\left|x^2-2\right|>x+2\)............................. (1)
th1: \(x^2-2\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge\sqrt{2}\\x\le-\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\) (1) \(\Leftrightarrow x^2-2>x+2\Leftrightarrow x^2-x-4>0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2>\dfrac{17}{4}\Leftrightarrow\left[{}\begin{matrix}x\ge\sqrt{\dfrac{17}{4}}+\dfrac{1}{2}\\x\le-\sqrt{\dfrac{17}{4}}+\dfrac{1}{2}\end{matrix}\right.\) (tmđk)
th2: \(x^2-2< 0\Leftrightarrow-\sqrt{2}< x< \sqrt{2}\)
\(\Rightarrow\) (1) \(\Leftrightarrow-x^2+2>x+2\Leftrightarrow-x^2>x\Leftrightarrow x^2+x< 0\)
\(\Leftrightarrow x\left(x+1\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x+1>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\-1< x< 0\end{matrix}\right.\) (tmđk)
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