Đáp án:
$\begin{array}{l}
A = 1\\
B = \dfrac{{x - 1}}{x}\left( {dkxd:x > 0;x \ne 1} \right)\\
B < A\\
\Rightarrow \dfrac{{x - 1}}{x} < 1\\
\Rightarrow \dfrac{{x - 1}}{x} - 1 < 0\\
\Rightarrow \dfrac{{x - 1 - x}}{x} < 0\\
\Rightarrow \dfrac{{ - 1}}{x} < 0\\
\Rightarrow x > 0\\
Vay\,x > 0;x \ne 1\\
2.1\\
M \in y = - 2x + 1\\
\Rightarrow M\left( {x; - 2x + 1} \right)\\
{d_{M - Ox}} = \left| {{y_M}} \right| = \left| { - 2x + 1} \right| = 2\\
\Rightarrow \left[ \begin{array}{l}
- 2x + 1 = 2\\
- 2x + 1 = - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - \dfrac{1}{2}\\
x = \dfrac{3}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
M\left( { - \dfrac{1}{2};2} \right)\\
M\left( {\dfrac{3}{2}; - 2} \right)
\end{array} \right.
\end{array}$
