Ta có:\(\frac{3^8.11+3^{10}.5}{3^9.2^4}\)=\(\frac{3^8.11+3^8.3^2.5}{3^9.2^4}\)=\(\frac{3^8.11+3^8.9.5}{3^9.2^4}\)=\(\frac{3^8.11+3^8.45}{3^9.2^4}\)=\(\frac{3^8.\left(11+45\right)}{3^9.2^4}\)=\(\frac{3^8.56}{3^9.2^4}\)=\(\frac{3^8.2^3.7}{3^9.2^4}\)=\(\frac{7}{3.2}\)=\(\frac{7}{6}\)

Để:A đạt giá trị nguyên

thì: \(3x+5⋮2+x\)và 2+x\(\ne0\)

\(\Leftrightarrow\)3x+6-1\(⋮\)2+x và x\(\ne-2\)

\(\Leftrightarrow\)3(x+2)-1\(⋮\)2+x

\(\Leftrightarrow-1⋮2+x\)

\(\Rightarrow\)2+x\(\in\)\(\cup\)(-1)=\(\left\{-1;1\right\}\)

Lập bảng:

\(\Rightarrow\)x\(\in\left\{0;2\right\}\)(TMĐK x\(\ne-2\))

Vậy: x\(\in\left\{0;2\right\}\)