Ta có: \(cos^2x+sin^2x=1\Rightarrow cos^2x=1-sin^2x\\ \Rightarrow cosx=\mp\sqrt{1-sin^2x}\)

Với \(sinx=\dfrac{2}{3}\Rightarrow cosx=\mp\sqrt{1-\dfrac{4}{9}}=\mp\sqrt{\dfrac{5}{9}}=\mp\dfrac{\sqrt{5}}{3}\)

Xét \(cosx=\dfrac{\sqrt{5}}{3}\) ta có:

\(A=18cos^2+9sin^2x-3cosx+6sinx\\ =18.\left(\dfrac{\sqrt{5}}{3}\right)^2+9.\left(\dfrac{2}{3}\right)^2-3.\dfrac{\sqrt{5}}{3}+6.\dfrac{2}{3}\\ =18.\dfrac{5}{9}+9.\dfrac{4}{9}-\sqrt{5}+4\\ =10+4-\sqrt{5}+4=18-\sqrt{5}\)

Với \(cosx=-\dfrac{\sqrt{5}}{3}\) làm tương tự.

Đs....

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

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

   \(\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\) ≤ \(\dfrac{1}{2}\) 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 \(\Rightarrow0\le x\le64\)