a)-2x-11\(⋮\)3x+2

\(\Rightarrow\)-3(-2x-11)\(⋮\)3x+2

\(\Rightarrow\)6x+33\(⋮\)3x+2

\(\Rightarrow\)(6x+4)+29\(⋮\)3x+2

\(\Rightarrow\)2(3x+4)+29\(⋮\)3x+2

\(\Rightarrow\)29\(⋮\)3x+2

\(\Rightarrow\)3x+2 \(\in\) Ư(29)={\(\pm1;\pm29\)}

nếu 3x+2=1 thì x=\(\frac{-1}{3}\)

nếu 3x+2=-1 thì x=-1

nếu 3x+2=29 thì x=9

nếu 3x+2=-29 thì x=\(\frac{-31}{3}\)

Vậy x \(\in\) {\(\frac{-1}{3};-1;9;\frac{-31}{3}\)}

b) x-7\(⋮\)x+6

\(\Rightarrow\)x+6-13\(⋮\)x+6

\(\Rightarrow\)13\(⋮\)x+6

\(\Rightarrow\)x+6 \(\in\) Ư(13)

Ta có Ư(13)={\(\pm1;\pm13\)}

nếu x+6=1 thì x=-5

nếu x+6=-1 thì x=-7

nếu x+6=13 thì x=7

nếu x+6=-13 thì x=-19

Vậy x \(\in\) {-5;-7;7;-19}