[THPT 2023] Phương trình \({{2}^{2{{x}^{2}}}}-{{6.2}^{{{x}^{2}}+x}}+{{2}^{2x+3}}=0\) có 4 nghiệm \({{x}_{1}}

\({{2}^{2{{x}^{2}}}}-{{6.2}^{{{x}^{2}}+x}}+{{2}^{2x+3}}=0\Leftrightarrow {{2}^{2{{x}^{2}}}}-{{6.2}^{{{x}^{2}}+x}}+{{8.2}^{2x}}=0\Leftrightarrow \frac{{{2}^{2{{x}^{2}}}}-{{6.2}^{{{x}^{2}}+x}}+{{8.2}^{2x}}}{{{2}^{2x}}}=0\Leftrightarrow {{2}^{2{{x}^{2}}-2x}}-{{6.2}^{{{x}^{2}}-x}}+8=0\)

Đặt \({{2}^{{{x}^{2}}-x}}=t,\,\,(t>0)\)

Phương trình trở thành  

\({t^2} – 6t + 8 = 0 \Leftrightarrow \left[ \begin{array}{l}
t = 2\\
t = 4
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
{2^{{x^2} – x}} = 2\\
{2^{{x^2} – x}} = 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{x^2} – x = 1\\
{x^2} – x = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{1 \pm \sqrt 5 }}{2}\\
x = – 1\\
x = 2
\end{array} \right.\)

Vì  \({{x}_{1}}<{{x}_{2}}<{{x}_{3}}<{{x}_{4}}\) nên \({{x}_{1}}=-1,\,\,{{x}_{2}}=\frac{1-\sqrt{5}}{2},\,\,{{x}_{3}}=\frac{1+\sqrt{5}}{2},\,\,{{x}_{4}}=2\)

\({{x}_{1}}+{{x}_{2}}+2{{x}_{3}}+{{x}_{4}}=(-1)+\frac{1-\sqrt{5}}{2}+2.\frac{1+\sqrt{5}}{2}+2=\frac{1}{2}\left( 5+\sqrt{5} \right)=\frac{1}{c}\left( a+\sqrt{b} \right),\,\,(a,b,c\in {{\mathbb{Z}}^{+}})\)

 \(\Rightarrow a=2,\,\,b=c=5\Rightarrow a.b.c=50\).

Chọn: A