d) \(u_1=\dfrac{3^1.\sqrt{1}}{2^1}=\dfrac{3}{2}\); \(u_2=\dfrac{3^2.\sqrt{2}}{2^2}=\dfrac{9\sqrt{2}}{4}\); \(u_3=\dfrac{3^3.\sqrt{3}}{2^3}=\dfrac{27\sqrt{2}}{8}\); \(u_4=\dfrac{3^4.\sqrt{4}}{2^4}=\dfrac{81}{8}\); \(u_5=\dfrac{3^5.\sqrt{5}}{2^5}=\dfrac{243\sqrt{2}}{32}\). Xét: \(\dfrac{u_n}{u_{n-1}}=\dfrac{3^n\sqrt{n}}{2^n}:\dfrac{3^{n-1}\sqrt{n-1}}{2^{n-1}}\)\(=\left(\dfrac{3}{2}\right)^{n-\left(n-1\right)}.\dfrac{\sqrt{n}}{\sqrt{n-1}}\) \(=\dfrac{3}{2}\sqrt{\dfrac{n}{n-1}}>\dfrac{3}{2}\sqrt{\dfrac{n}{n+n}}=\dfrac{3}{2}\sqrt{\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{4}>1\). Dễ thấy \(\left(u_n\right)\) là dãy số không âm nên \(u_n>u_{n-1}\). Vậy \(\left(u_n\right)\) là dãy số tăng.