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a) ĐK: x ≥ 0; x ≠ 9; x≠4

P= \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{3}{x-5\sqrt{x}+6}\right):\left(\dfrac{x+2}{\sqrt{x}-3}-\dfrac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)

= \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{x+2}{\sqrt{x}-3}-\dfrac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)

=\(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\dfrac{\left(x+2\right)\left(x-2\right)-x^2+\sqrt{x}+6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\)

=\(\dfrac{x-4+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\dfrac{x^2-4-x^2+\sqrt{x}+6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\)

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

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

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

=\(\dfrac{x^2-3x+2}{x-4}\)

b)  P ≤ -2

⇒ \(\dfrac{x^2-3x+2}{x-4}\) ≤ -2

⇔ \(\dfrac{x^2-3x+2}{x-4}\) + 2 ≤ 0

⇔ \(\dfrac{x^2-3x+2+2\left(x-4\right)}{x-4}\) ≤ 0

⇔ \(\dfrac{x^2-3x+2+2x-8}{x-4}\) ≤ 0

⇔\(\dfrac{x^2-x-6}{x-4}\) ≤ 0

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-x-6\ge0\\x-4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-x-6\le0\\x-4>0\end{matrix}\right.\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x\le2\\3\le x< 4\end{matrix}\right.\)

Vậy.......

a: ĐKXĐ: x>=0; x<>9; x<>4; x<>2

Ta có: \(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{3}{x-5\sqrt{x}+6}\right):\left(\frac{x+2}{\sqrt{x}-3}-\frac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\frac{\left(x+2\right)\left(x-2\right)-x^2+\sqrt{x}+6}{\left(x-2\right)\cdot\left(\sqrt{x}-3\right)}\)

\(=\frac{x-4+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\frac{x^2-4-x^2+\sqrt{x}+6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\cdot\frac{\left(x-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}+2}=\frac{\left(x-1\right)\left(x-2\right)}{x-4}\)

b: P<=-2

=>P+2<=0

=>\(\frac{\left(x-1\right)\left(x-2\right)+2\left(x-4\right)}{x-4}\le0\)

=>\(\frac{x^2-3x+2+2x-8}{x-4}\le0\)

=>\(\frac{x^2-x-6}{x-4}\le0\)

=>\(\frac{\left(x-3\right)\left(x+2\right)}{x-4}\le0\)

=>\(\frac{x-3}{x-4}\le0\)

=>3<=x<4

a) Ta có: \(P=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)

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

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

\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x-1}{x+\sqrt{x}+1}\)

\(=\dfrac{1}{x+\sqrt{x}+1}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{1}{x-1}\)

b) Thay \(x=5+2\sqrt{3}\) vào biểu thức \(P=\dfrac{1}{x-1}\), ta được:

\(P=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}\)

\(\Leftrightarrow P=\left(\dfrac{1}{\sqrt{3}+1}\right)^2\)

hay \(\sqrt{P}=\dfrac{\sqrt{3}-1}{2}\)

Vậy: Khi \(x=5+2\sqrt{3}\) thì \(\sqrt{P}=\dfrac{\sqrt{3}-1}{2}\)

a: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};-\dfrac{3}{2}\right\}\)