1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)

2:

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

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

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

b: \(2P=2\sqrt{x}+5\)

=>\(P=\sqrt{x}+\dfrac{5}{2}\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)

=>\(\sqrt{x}\left(2\sqrt{x}+5\right)=2\sqrt{x}+2\)

=>\(2x+3\sqrt{x}-2=0\)

=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)

=>\(2\sqrt{x}-1=0\)

=>x=1/4

Bài 1 : 

+) ĐKXĐ  : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

a) Ta có : 

\(x=4-2\sqrt{3}\)

\(\Leftrightarrow x=3-2\sqrt{3}+1\)

\(\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)( Thỏa mãn ĐKXĐ ) 

Vậy tại \(x=\left(\sqrt{3}-1\right)^2\)thì giá trị của biểu thức A là : 

\(A=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+1}{\sqrt{\left(\sqrt{3}-1\right)^2}-3}=\frac{\sqrt{3}-1+1}{\sqrt{3}-1-3}=\frac{\sqrt{3}}{\sqrt{3}-4}=\frac{-\sqrt{3}\left(\sqrt{3}+4\right)}{7}\)

b) 

\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\)

\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(B=\frac{-3-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

Ta có :

\(P=A:B\)

\(\Leftrightarrow P=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{-3\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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

c) \(P=\frac{-\sqrt{x}-3}{3}\ge0\)

Dấu bằng xảy ra 

\(\Leftrightarrow-\sqrt{x}-3=0\)

\(\Leftrightarrow\sqrt{x}=-3\)( vô lí )

Vậy không tìm được giá trị nào của x để P đạt GTNN

a: \(A=\dfrac{x\sqrt{x}+1}{x+2\sqrt{x}+1}\)

ĐKXĐ: x>=0

\(A=\dfrac{x\sqrt{x}+1}{x+2\sqrt{x}+1}\)

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

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

Thay x=4 vào A, ta được:

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

b: M=A*B

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\left(\dfrac{2x+6\sqrt{x}+7}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\left(\dfrac{2x+6\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right)\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{2x+6\sqrt{x}+7-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)^2}=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)

Để M>2 thì M-2>0

=>\(\dfrac{\sqrt{x}+6-2\sqrt{x}-2}{\sqrt{x}+1}>0\)

=>\(-\sqrt{x}+4>0\)

=>\(-\sqrt{x}>-4\)

=>\(\sqrt{x}< 4\)

=>0<=x<16

c: Để M là số nguyên thì \(\sqrt{x}+6⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1+5⋮\sqrt{x}+1\)

=>\(5⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\{1;-1;5;-5\right\}\)

=>\(\sqrt{x}\in\left\{0;-2;4;-6\right\}\)

=>\(\sqrt{x}\in\left\{0;4\right\}\)

=>\(x\in\left\{0;16\right\}\)