$\begin{array}{l}
1)y = \frac{x}{{{x^2} + x + 3 - m}}\\
HS\,xd\,tren\,R \Leftrightarrow {x^2} + x + 3 - m \ne 0,\forall x \in R\\
\Leftrightarrow pt\,{x^2} + x + 3 - m = 0\,vo\,nghiem\\
\Leftrightarrow \Delta = 1 - 4\left( {3 - m} \right) < 0 \Leftrightarrow - 11 + 4m < 0 \Leftrightarrow m < \frac{{11}}{4}\\
2)a)y = 2{x^3} + x = f\left( x \right)\\
Voi\,{x_1},{x_2} \in R\,ta\,co:\\
\frac{{f\left( {{x_1}} \right) - f\left( {{x_2}} \right)}}{{{x_1} - {x_2}}} = \frac{{2x_1^3 + {x_1} - 2x_2^3 - {x_2}}}{{{x_1} - {x_2}}} = \frac{{2\left( {{x_1} - {x_2}} \right)\left( {x_1^2 + {x_1}{x_2} + x_2^2} \right) + \left( {{x_1} - {x_2}} \right)}}{{{x_1} - {x_2}}}\\
= \frac{{\left( {{x_1} - {x_2}} \right)\left( {x_1^2 + {x_1}{x_2} + x_2^2 + 1} \right)}}{{{x_1} - {x_2}}} = x_1^2 + {x_1}{x_2} + x_2^2 + 1 > 0\\
Vay\,hsdong\,bien\,tren\,R.
\end{array}$
Ys b bạn xét tương tự nhé!
