\(\dfrac{2x+2}{\sqrt{x}}+\dfrac{\sqrt{x}^3-1^3}{x-\sqrt{x}}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1^3\right)}{x\left(\sqrt{x}+1\right)}\\ =\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{2x+2+x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\left(x>0;x\ne1\right)\)

   \(\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\) ≤ \(\dfrac{1}{2}\) Đk XĐ : x ≥ 0 

⇔ 1  - \(\dfrac{5}{\sqrt{x}+2}\) ≤ \(\dfrac{1}{2}\)

⇔        \(\dfrac{5}{\sqrt{x}+2}\) ≤ 1 - \(\dfrac{1}{2}\)

  ⇔      \(\dfrac{5}{\sqrt{x}+2}\) ≥  \(\dfrac{1}{2}\)

⇔       \(\dfrac{5}{\sqrt{x}+2}\) ≥ \(\dfrac{5}{10}\) 

⇔ \(\sqrt{x}\) + 2 ≤ 10

 ⇔ \(\sqrt{x}\) ≤ 8 

⇔ x ≤ 64

kết hợp với đk ta có:

0≤ x ≤ 64

Do \(\sqrt{x}+2>0;\forall x\ge0\) nên BPT tương đương:

\(2\left(\sqrt{x}-3\right)\le\sqrt{x}+2\)

\(\Leftrightarrow2\sqrt{x}-6\le\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}\le8\)

\(\Rightarrow x\le64\)

Kết hợp ĐKXĐ \(\Rightarrow0\le x\le64\)

   \(\dfrac{\sqrt{22}}{\sqrt{11}}\) + \(\sqrt{8}\) - \(\dfrac{7.\sqrt{2}}{3-\sqrt{2}}\)

= \(\sqrt{\dfrac{22}{11}}\) +  2\(\sqrt{2}\) - \(\dfrac{7.\sqrt{2}.(3+\sqrt{2})}{9-2}\)

= \(\sqrt{2}\) + 2\(\sqrt{2}\) - 3.\(\sqrt{2}\) - 2

= 3\(\sqrt{2}\)- 3\(\sqrt{2}\) - 2

= - 2