Cho hàm số $f\left( x \right)=\left( x+2a \right)\left( x+2b-a...

Ta có: ${f}'\left( x \right)=3a{{x}^{2}}\left( 2{{a}^{2}}+4ab+2 \right)x-2{{a}^{3}}+4{{a}^{2}}b+a+2b$.
Theo bài ra ta có: ${f}'\left( x \right)\ge 0$, $\forall x\in \mathbb{R}$.
$\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& a=0 \\
& 2x+2b\ge 0,x\in \mathbb{R} \\
\end{aligned} \right.không\ thỏa\ mãn \\
& \left\{ \begin{aligned}
& a>0 \\
& {\Delta }'\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a>0 \\
& {{\left( {{a}^{2}}+2ab+1 \right)}^{2}}-3a\left( -2{{a}^{3}}+4{{a}^{2}}b+a+2b \right)\le 0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& a>0 \\
& 7{{a}^{4}}-8{{a}^{3}}b+4{{a}^{2}}{{b}^{2}}-{{a}^{2}}-2ab+1\le 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a>0 \\
& {{\left( 2ab-\dfrac{4{{a}^{2}}+1}{2} \right)}^{2}}+\dfrac{3}{4}{{\left( 2{{a}^{2}}-1 \right)}^{2}}\le 0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a>0 \\
& 2ab-\dfrac{4{{a}^{2}}+1}{2}=0 \\
& 2{{a}^{2}}-1=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{\sqrt{2}}{2} \\
& b=\dfrac{3\sqrt{2}}{4} \\
\end{aligned} \right.$
Nhận xét: Hàm số $f\left( x \right)=k\left( x+a \right)\left( x+b \right)\left( x+c \right)$ đồng biến trên $\mathbb{R}$ khi $\left\{ \begin{aligned}
& k>0 \\
& a=b=c \\
\end{aligned} \right. $; nghịch biến trên $ \mathbb{R} $ khi $ \left\{ \begin{aligned}
& k<0 \\
& a=b=c \\
\end{aligned} \right.$