$\displaystyle \begin{array}{{>{\displaystyle}l}} B=\left(\frac{1}{\sqrt{x} +2} +\frac{1}{\sqrt{x} -2}\right) .\frac{\sqrt{x} -2}{\sqrt{x}} \ \\ DK:x\#4\\ B=\frac{\sqrt{x} -2+\sqrt{x} +2}{x-4} .\frac{\sqrt{x} -2}{\sqrt{x}} \ \\ B=\frac{2\sqrt{x}}{\left(\sqrt{x} +2\right)\sqrt{x}} =\frac{2}{\sqrt{x} +2} \ \\ b) \ B >\frac{1}{2} \ \\ \rightarrow \frac{2}{\sqrt{x} +2} -\frac{1}{2} \ >0\ \\ \rightarrow \frac{4-\sqrt{x} -2}{2\sqrt{x} +4} \ \\ \rightarrow \frac{2-\sqrt{x}}{2\sqrt{x} +4} \ \\ Nhận\ thấy\ 2\sqrt{x} +4\ >0\ với\ mọi\ x\ \\ Do\ đó\ :\ Để\ B-\frac{1}{2} \ thì\ 2-\sqrt{x} \ >0\ \\ \rightarrow \sqrt{x} < 2\\ \rightarrow x< 4\ \\ Vậy\ để.......\ \\ c) A=\frac{14}{3\sqrt{x} +6} \ nguyên\ \\ \rightarrow Thì\ 3\sqrt{x} +6\ =\frac{14}{a} \ ( a\ thuộc\ Z;a\#0) \ \\ \rightarrow \ \sqrt{x} =\frac{\frac{14}{a} -6}{3} =\frac{14-6a}{3a} \ \\ \rightarrow x=\left(\frac{14-6a}{3a}\right)^{2}( x\geqslant 0) \ \\ Để\ x\#0\ \ \frac{14-6a}{3a} \#0\ \\ \rightarrow 14-6a\#0\ \rightarrow a\#\frac{7}{3}\\ Để\ x\#4\ \rightarrow \left(\frac{14-6a}{3a}\right)^{2} -4\#0\ \\ \rightarrow \left(\frac{14-6a}{3a} -2\right)\left(\frac{14-6a}{3a} +2\right) \#0\ \\ \rightarrow \ \left(\frac{14-6a-6a}{3a}\right)\left(\frac{14-6a+6a}{3a}\right) \#0\ \\ \rightarrow 14-12a\#0\ \rightarrow a=\frac{14}{12} =\frac{7}{6} \ \\ Vậy\ với\ a\ thuộc\ Z\ a\ khác\ \frac{7}{6} \ và\ \frac{7}{3} \ \\ thì\ x=\left(\frac{14-6a}{3a}\right)^{2} \ làm\ cho\ A\ nguyên\ \\ \\ \\ \\ \\ \\ \\ \end{array}$
