f) $y= f(x)= \sqrt{x^2 + 1}$
$TXD: D = \Bbb R$
Chọn $x_2, \, x_1 \in \Bbb R$
Xét $\dfrac{f(x_2) - f(x_1)}{x_2-x_1}$
$= \dfrac{\sqrt{x_2^2 +1} - \sqrt{x_1^2 + 1}}{x_2 - x_1}$
$= \dfrac{(\sqrt{x_2^2 +1} - \sqrt{x_1^2 + 1})(\sqrt{x_2^2 +1} +\sqrt{x_1^2 + 1}}{(x_2-x_1)(\sqrt{x_2^2 +1} + \sqrt{x_1^2 + 1})}$
$= \dfrac{x_2^2 + 1 - (x_1^2 +1)}{(x_2-x_1)(\sqrt{x_2^2 +1} + \sqrt{x_1^2 + 1})}$
$= \dfrac{1}{\sqrt{x_2^2 +1} +\sqrt{x_1^2 + 1}}$
Ta có:
$\sqrt{x_2^2 +1} +\sqrt{x_1^2 + 1} > 0, \, \forall x$
$\to \dfrac{1}{\sqrt{x_2^2 +1} +\sqrt{x_1^2 + 1}} > 0$
$\to \dfrac{f(x_2) - f(x_1)}{x_2-x_1} >0$
Vậy hàm số đồng biến trên $\Bbb R$
k) $y = f(x) = \sqrt{2x +1}$
$TXD: D = \left[-\dfrac{1}{2};+\infty\right)$
Chọn $x_1, x_2\in D \quad (x_1\ne x_2)$
Xét $\dfrac{f(x_2) - f(x_1)}{x_2-x_1}$
$= \dfrac{\sqrt{2x_2 +1} - \sqrt{2x_1+1}}{x_2 - x_1}$
$= \dfrac{(\sqrt{2x_2 +1} - \sqrt{2x_1+1})(\sqrt{2x_2 +1} +\sqrt{2x_1+1})}{(x_2-x_1)(\sqrt{2x_2 +1} + \sqrt{2x_1+1})}$
$= \dfrac{2x_2 + 1 - (2x_1 +1)}{(x_2 - x_1)(\sqrt{2x_2 +1} + \sqrt{2x_1+1})}$
$= \dfrac{2}{\sqrt{2x_2 +1} +\sqrt{2x_1+1}}$
Ta có:
$\sqrt{2x_2 +1} +\sqrt{2x_1+1} > 0, \, \forall x \in D$
$\to \dfrac{2}{\sqrt{2x_2 +1} +\sqrt{2x_1+1}} > 0$
$\to \dfrac{f(x_2) - f(x_1)}{x_2-x_1} > 0$
Vậy hàm số đồng biến trên $\left[-\dfrac{1}{2};+\infty\right)$
