Đáp án: 1.B, 2B
Giải thích các bước giải:
$\begin{array}{l}
1)\left| {\left( {{x^2} - x} \right)} \right| \le {x^2} - x\\
+ Khi:{x^2} - x \ge 0 \Rightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le 0
\end{array} \right.\\
\Rightarrow {x^2} - x \le {x^2} - x\left( {luon\,dung} \right)\\
+ Khi:{x^2} - x \le 0 \Rightarrow 0 \le x \le 1\\
\Rightarrow - {x^2} + x \le {x^2} - x\\
\Rightarrow {x^2} - x \ge 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
Vậy\,x \ge 1\,hoặc\,x \le 0\\
\Rightarrow B\\
2)\\
m > 2x + \dfrac{1}{{{x^2}}}\,khi\,x > 0\\
\Rightarrow Theo\,Cô - si:\\
2x + \dfrac{1}{{{x^2}}} = x + x + \dfrac{1}{{{x^2}}} \ge 3\sqrt[3]{{x.x.\dfrac{1}{{{x^2}}}}} = 3\\
\Rightarrow 2x + \dfrac{1}{{{x^2}}} \ge 3\\
\Rightarrow m > 2x + \dfrac{1}{{{x^2}}} \ge 3\\
\Rightarrow m > 3\\
\Rightarrow B
\end{array}$
