\(P=A.B=\dfrac{\sqrt{x}}{\sqrt{x}+1}.\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

Ta có : \(\left|P\right|-P=0\) \(\Leftrightarrow\left|P\right|=P\Leftrightarrow\left|\dfrac{\sqrt{x}}{\sqrt{x}-2}\right|=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(+TH_1:x\ge0\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\) (luôn đúng)

\(+TH_2:x< 0\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}=0\)

\(\Leftrightarrow-2.\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)=0\)

\(\Leftrightarrow x=0\)

\(A=P:Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}:\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+4}=1+\dfrac{-5}{\sqrt{x}+4}\)

Điều kiện : \(x\ge4\Rightarrow\sqrt{x}+4\ge4\Rightarrow-\dfrac{5}{\sqrt{x}+4}\le-\dfrac{5}{4}\Rightarrow\dfrac{5}{\sqrt{x}+4}\ge\dfrac{5}{4}\)

Dấu ''='' xảy ra \(\Leftrightarrow x=0\)

Vậy \(min_A=\dfrac{5}{4}\Leftrightarrow x=0\)

\(P=A.B=\dfrac{2\sqrt{x}}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)

Ta có : \(\sqrt{P}\le\dfrac{\sqrt{5}}{2}\Rightarrow\sqrt{\dfrac{2\sqrt{x}}{\sqrt{x}+1}}\le\dfrac{\sqrt{5}}{2}\left(dkxd:x\ge0\right)\)

Bình phương 2 vế bất pt, ta được :

\(\dfrac{2\sqrt{x}}{\sqrt{x}+1}\le\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{2.4\sqrt{x}-5\left(\sqrt{x}+1\right)}{4\left(\sqrt{x}+1\right)}\le0\)

\(\Leftrightarrow8\sqrt{x}-5\sqrt{x}-5\le0\)

\(\Leftrightarrow3\sqrt{x}\le5\)

\(\Leftrightarrow\sqrt{x}\le\dfrac{5}{3}\)

\(\Leftrightarrow x\le\dfrac{25}{9}\)

Mà x phải là giá trị nguyên nên \(x\le2\) (với \(x\in Z\))

So với điều kiện \(x\ge0\Rightarrow0\le x\le2\)

Vậy \(x\in\left\{0;1;2\right\}\)

c,M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) :  \(\dfrac{\sqrt{x}+3}{\sqrt{x}+5}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) \(\times\) \(\dfrac{\sqrt{x}+5}{\sqrt{x}+3}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\) = \(\dfrac{\sqrt{x}+3-7}{\sqrt{x}+3}\)

 M = 1  - \(\dfrac{7}{\sqrt{x}+3}\) 

 M \(\in\) Z ⇔ 7 ⋮ \(\sqrt{x}\) + 3 vì \(\sqrt{x}\) ≥ 0 ⇒ \(\sqrt{x}\) + 3 ≥ 3 ⇒ 0< \(\dfrac{7}{\sqrt{x}+3}\) ≤ \(\dfrac{7}{3}\)

⇒ M Đạt giá trị nguyên lớn nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) đạt giá trị nguyên nhỏ nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) = 1 ⇔ \(\sqrt{x}\) + 3  = 7 ⇔ \(\sqrt{x}\) = 4 ⇔ \(x\) = 16 

Mnguyên(max)  = 1 - 1 = 0 xảy ra khi \(x\) = 16

Ta có :

\(A.B=\dfrac{24}{\sqrt{x}+6}.\dfrac{\sqrt{x}+6}{\sqrt{x}-6}\)

\(=\dfrac{24}{\sqrt{x}-6}\)

Để \(AB\le12\Leftrightarrow\dfrac{24}{\sqrt{x}-6}\le12\)

\(\Leftrightarrow\dfrac{24-12\left(\sqrt{x}-6\right)}{\sqrt{x}-6}\le0\)

\(\Leftrightarrow24-12\sqrt{x}+72\le0\)

\(\Leftrightarrow-12\sqrt{x}\le-96\)

\(\Leftrightarrow\sqrt{x}\ge8\)

\(\Leftrightarrow x\ge64\)

Vậy \(x\ge64\) thì \(AB\le12\)

\(P=A:B=\dfrac{1-\sqrt{x}}{\sqrt{x}-2}:\dfrac{2\sqrt{x}}{\sqrt{x}-2}=\dfrac{1-\sqrt{x}}{2\sqrt{x}}\)

Có: \(\left|P+1\right|< 3P\left(ĐK:x>0\right)\)

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

Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\) nên:

\(\left|\dfrac{\sqrt{x}+1}{2\sqrt{x}}\right|< \dfrac{3-3\sqrt{x}}{2\sqrt{x}}\\ \Leftrightarrow\dfrac{\sqrt{x}+1-3+3\sqrt{x}}{2\sqrt{x}}< 0\\ \Leftrightarrow\dfrac{4\sqrt{x}-2}{2\sqrt{x}}< 0\\ \Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}}< 0\\ \Rightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\2\sqrt{x}-1< 0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow0< x< \dfrac{1}{4}\)

c: P nguyên

=>căn x+1+4 chia hết cho căn x+1

=>căn x+1 thuộc {1;2;4}

=>x thuộc {1;9}

a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

b) Ta có: \(x=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{2}+1-\sqrt{2}+1\)

=2

Thay x=2 vào A, ta được:

\(A=\dfrac{-3}{3+\sqrt{2}}=\dfrac{-9+3\sqrt{2}}{7}\)