a) Ta có: \(B=\dfrac{3}{2}\)
nên \(\dfrac{6n+3}{6n-2}=\dfrac{3}{2}\)
\(\Leftrightarrow18n-6=12n+6\)
\(\Leftrightarrow18n-12n=6+6\)
\(\Leftrightarrow6n=12\)
hay n=2(thỏa ĐK)
b) Để B nguyên thì \(6n+3⋮6n-2\)
\(\Leftrightarrow5⋮6n-2\)
\(\Leftrightarrow6n-2\in\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow6n\in\left\{3;1;7;-3\right\}\)
\(\Leftrightarrow n\in\left\{\dfrac{1}{2};\dfrac{1}{6};\dfrac{7}{6};-\dfrac{1}{2}\right\}\)(loại)