Cho bất phương trình ${{\log }_{7}}\left( {{x}^{2}}+2x+2...

$bpt\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}+6x+5+m>0 \\
& {{\log }_{7}}\left[ 7\left( {{x}^{2}}+2x+2 \right) \right]>{{\log }_{7}}\left( {{x}^{2}}+6x+5+m \right) \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& m>-{{x}^{2}}-6x-5 \\
& 6{{x}^{2}}+8x+9>m \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m>\underset{\left( 1; 3 \right)}{\mathop{\max }} f\left( x \right) \\
& m<\underset{\left( 1; 3 \right)}{\mathop{\min }} g\left( x \right) \\
\end{aligned} \right. f\left( x \right)=-{{x}^{2}}-6x-5 g\left( x \right)=6{{x}^{2}}+8x+9f\left( x \right)g\left( x \right){f}'\left( x \right)=-2x-6<0, \forall x\in \left( 1; 3 \right)\Rightarrow f\left( x \right)\left( 1; 3 \right)\Rightarrow \underset{\left( 1; 3 \right)}{\mathop{\max }} f\left( x \right)=f\left( 1 \right)=-12{g}'\left( x \right)=12x+8>0, \forall x\in \left( 1; 3 \right)\Rightarrow g\left( x \right)\left( 1; 3 \right)\Rightarrow \underset{\left( 1; 3 \right)}{\mathop{\min }} g\left( x \right)=g\left( 1 \right)=23-12<m<23m\in \mathbb{Z}m\in \left\{ -11; -10; ...; 22 \right\}34m$ thỏa mãn yêu cầu bài toán.