Mai: I didn’t get the job I applied for.

                        Tom: _____________ You’ll find something better soon.

Tam thức bậc hai \(f\left( x \right) = {x^2} + 2mx + m - 2\) có:

\(\Delta ' = {m^2} - m + 2 = {\left( {m - \frac{1}{2}} \right)^2} + \frac{7}{4} \ge \frac{7}{4} > 0\) với mọi \(m \in \mathbb{R}\).

Do đó \(f\left( x \right)\) có hai nghiệm phân biệt \({x_1},{x_2}\left( {{x_1} < {x_2}} \right)\) với mọi \(m \in \mathbb{R}\).

Để \(f\left( x \right) < 0\) với mọi \(x \in \left( {1;\,\,2} \right)\) thì \({x_1} \le 1 < 2 \le {x_2}\)

\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{f\left( 1 \right) \le 0}\\{f\left( 2 \right) \le 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{1 + 2m + m - 2 \le 0}\\{4 + 4m + m - 2 \le 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{3m - 1 \le 0}\\{5m + 2 \le 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{m \le \frac{1}{3}}\\{m \le \frac{{ - 2}}{5}}\end{array}} \right. \Leftrightarrow m \le \frac{{ - 2}}{5}\].

Vậy \(m \le \frac{{ - 2}}{5}\) thì \({x^2} + 2mx + m - 2 < 0\) với mọi \(x \in \left( {1;\,\,2} \right)\). Chọn D.