Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ đồng...

Ta có: $$
$\Leftrightarrow {{f}^{2}}\left( x \right)+6x.f\left( x \right)+9{{\text{x}}^{2}}=9{{\text{x}}^{4}}+12{{\text{x}}^{2}}+4\Leftrightarrow {{\left[ f\left( x \right)+3\text{x} \right]}^{2}}={{\left( 3{{x}^{2}}+2 \right)}^{2}}\Leftrightarrow \left[ \begin{align}
& f\left( x \right)=3{{x}^{2}}-3x+2 \\
& f\left( x \right)=-3{{\text{x}}^{2}}-3x-2 \\
\end{align} \right. f\left( 0 \right)>0 f\left( x \right)=3{{x}^{2}}-3x+2y=f\left( 2{{\text{x}}^{2}}-3\text{x}+1 \right)2{{\text{x}}^{2}}-3x+1=tx\in \left[ 0;1 \right]\Rightarrow t\in \left[ -\dfrac{1}{8};1 \right]f\left( t \right)=3{{t}^{2}}-3t+2\left[ -\dfrac{1}{8};1 \right]{f}'\left( t \right)=6t-3{f}'\left( t \right)=0\Leftrightarrow t=\dfrac{1}{2}\left\{ \begin{aligned}
& f\left( -\dfrac{1}{8} \right)=\dfrac{155}{64} \\
& f\left( \dfrac{1}{2} \right)=\dfrac{5}{4} \\
& f\left( 1 \right)=2 \\
\end{aligned} \right. \underset{\left[ 0;1 \right]}{\mathop{\max }} f\left( 2{{\text{x}}^{2}}-3\text{x}+1 \right)=\dfrac{155}{64}$.