Đáp án:
$\begin{array}{l}a. \ \rm Tính \ giá \ trị:\\\text{Ta có:} \ x=36 \to \sqrt{x}=\sqrt{36}=6\\\Rightarrow B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\\b. \ \rm Rút \ gọn:\\A=\left({\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\\quad=\left({\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}}\right)\times\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\\\quad=\dfrac{x-1}{\sqrt{x}(\sqrt{x}-1)}\times\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\\\quad=\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}\times\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\\\quad=\dfrac{\sqrt{x}+2}{\sqrt{x}}\end{array}$
