Cho bất phương trình $m{{.3}^{x+1}}+(3m+2).{{\left( 4-\sqrt{7}...

Leftrightarrow {{\left( \frac{4+\sqrt{7}}{3} \right)}^{x}}+\left( 3m+2 \right){{\left( \frac{4-\sqrt{7}}{3} \right)}^{x}}+3m>0t={{\left( \frac{4+\sqrt{7}}{3} \right)}^{x}}\Rightarrow \frac{1}{t}={{\left( \frac{4-\sqrt{7}}{3} \right)}^{x}}.x\in (-\infty ;0]\Rightarrow t\in \left( 0;1 \right]{{t}^{2}}+3mt+\left( 3m+2 \right)>0mm \in(-2022 ; 2023){{t}^{2}}+3mt+\left( 3m+2 \right)>0t\in (0;1]{{t}^{2}}+3mt+\left( 3m+2 \right)>0\Leftrightarrow -3m<\frac{{{t}^{2}}+2}{t+1},\left( t\in \left( 0;1 \right] \right).h\left( t \right)=\frac{{{t}^{2}}+2}{t+1}\Rightarrow {h}'\left( t \right)=\frac{{{t}^{2}}+2t-2}{{{\left( t+1 \right)}^{2}}}\Rightarrow {h}'\left( t \right)=0\Leftrightarrow \left[ \begin{align}
& t=-1+\sqrt{3}\in \left( 0;1 \right] \\
& t=-1-\sqrt{3}\notin \left( 0;1 \right] \\
\end{align} \right. Vậy để bất phương trình luôn đúng với mọi \)">t\in (0;1]-3m<-2+4\sqrt{3}\Rightarrow m>\frac{2-4\sqrt{3}}{3}\approx -1.64.mm \in(-2022 ; 2023)2024$ giá trị thỏa mãn.