a) Ta có \( y'=\dfrac{(x^{2}+x+2)'.(x-1)-(x^{2}+x+2).(x-1)'}{(x-1)^{2}}\) = \( \dfrac{x^{2}-2x-3}{(x-1)^{2}}\)
Do đó, \(y'<0\Leftrightarrow \dfrac{x^{2}-2x-3}{(x-1)^{2}}<0\)
\( \Leftrightarrow \left\{ \matrix{
x \ne 1 \hfill \cr
- 1 < x < 3 \hfill \cr} \right. \Leftrightarrow \)\(x∈ (-1;1) ∪ (1;3)\).
b) Ta có \( y'=\dfrac{(x^{2}+3)'.(x+1)-(x^{2}+3).(x+1)'}{(x+1)^{2}}\) = \( \dfrac{x^{2}+2x-3}{(x+1)^{2}}\).
Do đó, \(y'≥0 \Leftrightarrow \dfrac{x^{2}+2x-3}{(x+1)^{2}}≥0 \)
\( \Leftrightarrow \left\{ \matrix{ x \ne - 1 \hfill \cr \left[ \matrix{ x \ge 1 \hfill \cr x \le - 3 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x \ge 1 \hfill \cr x \le - 3 \hfill \cr} \right. \)
\(\Leftrightarrow x∈ (-∞;-3] ∪ [1;+∞)\).
c) Ta có \( y'=\dfrac{(2x-1)'.(x^{2}+x+4)-(2x-1).(x^{2}+x+4)'}{(x^{2}+x+4)^2}\) \(=\dfrac{-2x^{2}+2x+9}{(x^{2}+x+4)^2}\).
Do đó, \(y'>0 \Leftrightarrow \dfrac{-2x^{2}+2x+9}{(x^{2}+x+4)^2} >0\Leftrightarrow -2x^2+2x +9>0 \)\(\Leftrightarrow \dfrac{1-\sqrt{19}}{2} < x < \dfrac{1+\sqrt{19}}{2}\Leftrightarrow x∈ \left ( \dfrac{1-\sqrt{19}}{2};\dfrac{1+\sqrt{19}}{2} \right )\)
Vì \(x^2+x +4 =\) \( \left ( x+\dfrac{1}{2} \right )^{2}\)+ \( \dfrac{15}{4} >0\), với \(∀ x ∈ \mathbb R\).
