Cho các số thực $a$, $b$ thỏa mãn...

Ta có: ${{\text{e}}^{{{a}^{2}}+2{{b}^{2}}}}+{{\text{e}}^{ab}}\left( {{a}^{2}}-ab+{{b}^{2}}-1 \right)-{{\text{e}}^{1+ab+{{b}^{2}}}}=0$ $\Leftrightarrow {{\text{e}}^{{{a}^{2}}-ab+2{{b}^{2}}}}+{{a}^{2}}-ab+{{b}^{2}}-1-{{\text{e}}^{1+{{b}^{2}}}}=0$
$\Leftrightarrow {{\text{e}}^{{{a}^{2}}-ab+2{{b}^{2}}}}+{{a}^{2}}-ab+2{{b}^{2}}={{\text{e}}^{1+{{b}^{2}}}}+1+{{b}^{2}}$, $\left( * \right)$
Xét hàm số: $f\left( t \right)={{\text{e}}^{t}}+t$ $\Rightarrow {f}'\left( t \right)={{\text{e}}^{t}}+1>0$ $\Rightarrow $ $f\left( t \right)={{\text{e}}^{t}}+t$ là hàm số đồng biến trên $\mathbb{R}$
$\left( * \right)\Leftrightarrow f\left( {{a}^{2}}-ab+2{{b}^{2}} \right)=f\left( 1+{{b}^{2}} \right)\Leftrightarrow {{a}^{2}}-ab+2{{b}^{2}}=1+{{b}^{2}}\Leftrightarrow {{a}^{2}}-ab+{{b}^{2}}=1$
Do đó: $P=\dfrac{1}{1+2ab}=\dfrac{{{a}^{2}}-ab+{{b}^{2}}}{{{a}^{2}}+ab+{{b}^{2}}}$
TH1: $b=0\Rightarrow P=1$
TH2: $b\ne 0\Rightarrow P=\dfrac{{{\left( \dfrac{a}{b} \right)}^{2}}-\left( \dfrac{a}{b} \right)+1}{{{\left( \dfrac{a}{b} \right)}^{2}}+\left( \dfrac{a}{b} \right)+1}=\dfrac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1}=g\left( x \right)$ với $x=\dfrac{a}{b}$
${g}'\left( x \right)=\dfrac{2{{x}^{2}}-2}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}$ ; ${g}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=-1 \\
\end{aligned} \right.$
Bảng biến thiên
Do đó: $m={{P}_{\min }}=\dfrac{1}{3}$ ; $M={{P}_{\max }}=3$ $\Rightarrow m+M=\dfrac{10}{3}$.