a) \(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

a, - Thay x = 25 vào biểu thức B ta được :

\(B=\frac{\sqrt{25}-3}{\sqrt{25}+1}=\frac{1}{3}\)

b, Ta có : \(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

=> ​​​\(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}\)​

=> \(A=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

=> \(A=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{x-9}\)

=> \(A=\frac{2x-6\sqrt{x}+x+\sqrt{x}+3\sqrt{x}+3+11\sqrt{x}-3}{x-9}\)

=> \(A=\frac{3x+9\sqrt{x}}{x-9}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

c, Ta có : \(P=AB+1=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+1\)

=> \(P=\frac{3\sqrt{x}}{\sqrt{x}+1}+1\)

=> \(P=\frac{3\sqrt{x}+3-3}{\sqrt{x}+1}+1\)

=> \(P=4-\frac{3}{\sqrt{x}+1}\)

Ta thấy : \(\sqrt{x}\ge0\)

=> \(4-\frac{3}{\sqrt{x}+1}\ge1\)

Vậy Min_P = 1 khi x = 0

a) \(A=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(A=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(A=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3}{\sqrt{x}+3}\)

b) \(A=\frac{1}{3}=>\frac{3}{\sqrt{x}+3}=\frac{1}{3}\)
\(=>\sqrt{x}+3=9\)

\(=>\sqrt{x}=6=>x=36\)
c) \(A\)\(lớn\)\(nhất\)\(< =>\frac{3}{\sqrt{x}+3}lớn\)\(nhất\)
\(=>\sqrt{x}+3\)\(nhỏ\)\(nhất\)
\(Mà\)\(\sqrt{x}+3>=3 \)
\(Do\)\(đó\)\(\sqrt{x}+3=3=>x=0\)

a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)

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

Thay x = 4 => \(\sqrt{x}=2\) vào B ta được : 

\(B=\frac{2+5}{2-3}=-7\)

b, Ta có : Với \(x\ge0;x\ne9\)

\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)

\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)

Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)

B đâu ra chỉ? Không biết đề có sai không chứ mình rút gọn ra nhiêu đây thì ko đủ chứng minh C\(\ge0\) được

a) \(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{4\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{4\sqrt{x}-12}{x-9}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{x+3\sqrt{x}}{x-9}\)

\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}\)

\(=\frac{x-25}{x-9}\)

b) \(P=\frac{A}{B}=\frac{\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}\)

\(=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)

\(\sqrt{P}< \frac{1}{3}\Rightarrow\sqrt{\frac{\sqrt{x}-5}{\sqrt{x}+3}}< \frac{1}{3}\)

\(\Rightarrow\frac{\sqrt{x}-5}{\sqrt{x}+3}< \frac{1}{9}\Leftrightarrow9\sqrt{x}-45< \sqrt{x}+3\)

\(\Leftrightarrow8\sqrt{x}< 48\Leftrightarrow\sqrt{x}< 6\Rightarrow0\le x< 36\)

\(a,\)\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(b,P=\frac{A}{B}=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+5}{\sqrt{x}-3}\)

\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}-3\right)}{\sqrt{x}+5}=\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}\)

Để \(\sqrt{P}< \frac{1}{3}\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}< \frac{1}{3}\)

\(\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}-\frac{1}{3}< 0\)

\(\Rightarrow\frac{3\left(2x+3\sqrt{x}-1\right)-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)

\(\Rightarrow6x+9\sqrt{x}-3-\sqrt{x}-5< 0\)( do \(3\left(\sqrt{x}+5\right)>0\))

\(\Rightarrow6x-8\sqrt{x}-8< 0\Rightarrow3x-4\sqrt{x}-4< 0\)

\(\Rightarrow3x-6\sqrt{x}+2\sqrt{x}-4< 0\)

\(\Rightarrow3\sqrt{x}\left(\sqrt{x}-2\right)+2\left(\sqrt{x}-2\right)< 0\)

\(\Rightarrow\left(\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)< 0\)

Vì \(3\sqrt{x}+2>0\Rightarrow\sqrt{x}-2< 0\)

\(\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

Vậy để \(\sqrt{P}< \frac{1}{3}\)thì \(0\le x< 4\)