a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne9\\x\ne4\end{cases}}\)

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

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

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

\(\Leftrightarrow P=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)

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

\(\Leftrightarrow P=\frac{4x}{\sqrt{x}-3}\)

b) Để P < 0

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-3< 0\Leftrightarrow4x>0\\\sqrt{x}-3>0\Leftrightarrow4x< 0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< 3\Leftrightarrow x>0\\\sqrt{x}>3\Leftrightarrow x< 0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x< 9\Leftrightarrow x>0\left(ktm\right)\\x>9\Leftrightarrow x< 0\left(ktm\right)\end{cases}}\)

Vậy để \(P< 0\Leftrightarrow x\in\varnothing\)

Để P > 0

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-3>0\Leftrightarrow4x>0\\\sqrt{x}-3< 0\Leftrightarrow4x< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}>3\Leftrightarrow x>0\left(tm\right)\\\sqrt{x}< 3\Leftrightarrow x< 0\left(ktm\right)\end{cases}}\)

\(\Leftrightarrow x>9\Leftrightarrow x>0\left(tm\right)\)

Vậy để \(P>0\Leftrightarrow x>9\)

c) Để  \(\left|P\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}P=1\left(tm\right)\\P=-1\left(ktm\right)\end{cases}}\)

\(\Leftrightarrow\frac{4x}{\sqrt{x}-3}=1\)

\(\Leftrightarrow4x=\sqrt{x}-3\)

\(\Leftrightarrow4x-\sqrt{x}+3=0\)

\(\Leftrightarrow\left(2\sqrt{x}-\frac{1}{4}\right)^2+\frac{47}{48}=0\left(ktm\right)\)

Vậy để \(\left|P\right|=1\Leftrightarrow x\in\varnothing\)

\(\sqrt{x^2+2x+1}+\sqrt{x^4-2x^2+2}=1\)

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

Mà \(\sqrt{\left(x+1\right)^2}+\sqrt{\left(x^2-1\right)^2+1}\ge1\)

nên dấu "=" <=> x = -1

\(\sqrt{x^2+2x+1}+\sqrt{x^4-2x^2+2}=1\)

<=> \(\sqrt{x^2+2x+1}=1-\sqrt{x^4-2x^2+2}\)

<=> \(\left(\sqrt{x^2+2x+1}\right)^2=\left(1-\sqrt{x^4-2x^2+2}\right)^2\)

<=> x² + 2x + 1 = x⁴ - 2x² + 3 - 2\(\sqrt{x^4-2x^2+2}\)

<=> x² + 2x + 1 - (x⁴ - 2x) = -2\(\sqrt{x^4-2x^2+2}\) - (x⁴ - 2x)

<=> -x⁴ + 3x² + 1 = -2\(\sqrt{x^4-2x^2+2}+3\)

<=> -x⁴ + 3x² + 1 - 3 = -2\(\sqrt{x^4-2x^2+2}\)

<=> (-x⁴ + 3x² - 2)² = (-2\(\sqrt{x^4-2x^2+2}\))²

<=> x⁸ - 6x⁶ - 4x⁵ + 13x⁴ + 12x³ - 8x² - 8x + 4 = 4x⁴ - 8x² + 8

<=> x = -1

=> x = -1