a) Đkxđ: x>/2
\(B=\left(\dfrac{\sqrt{x+2}}{\sqrt{x-1}}-\dfrac{\sqrt{x-2}}{\sqrt{x+1}}\right)\dfrac{x-1}{\sqrt{x+2}}=\sqrt{x-1}-\dfrac{\sqrt{x-2}\left(x-1\right)}{\sqrt{x+1}\cdot\sqrt{x+2}}\)
b) \(B\in Z\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}\in Z\\\dfrac{\sqrt{x-2}\left(x-1\right)}{\sqrt{x+1}\sqrt{x+2}}\in Z\end{matrix}\right.\)
+ Xét x=2, ta có: B= 1 (tm)
+Xét x > 2
Vì \(x\in Z\) nên \(x\ge3\)
với \(x\ge3\), ta có:
\(\left(x-2\right)\sqrt{x-1}\ge2\)
\(\sqrt{x+1}\cdot\sqrt{x+2}\ge2\sqrt{5}\)
vậy với \(x\ge3\) ta luôn có:
\(\left(x-2\right)\sqrt{x-1}< \)\(\sqrt{x+1}\cdot\sqrt{x+2}\)
hay \(\dfrac{\sqrt{x-2}\left(x-1\right)}{\sqrt{x+1}\cdot\sqrt{x+2}}\notin Z\) \(\Rightarrow B\notin Z\)
KL: a) x >/ 2
b) x=2
