\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)

=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

b/ Để R<-1   => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)

<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)

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

Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)   là sao vậy ạ?

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

ĐK  \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

a, \(R=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

b. \(R< -1\Rightarrow R+1< 0\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)

\(\Rightarrow0\le x< \frac{9}{4}\)

c. \(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)

Ta thấy \(\sqrt{x}+3\ge3\Rightarrow\frac{-18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\Rightarrow R\ge-3\)

Vậy \(MinR=-3\Leftrightarrow x=0\)

\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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

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

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

\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)

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

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

\(b)\) Ta có : \(R< -1\)

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

\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)

\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)

\(\Leftrightarrow\)\(4\sqrt{x}< 6\)

\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)

\(\Leftrightarrow\)\(x< \frac{9}{4}\)

Chúc bạn học tốt ~ 

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

   \(R=\left(\frac{\sqrt{x}\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\left(\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)

  \(R=\left(\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)

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

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

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

c

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

\(co:x>o\inĐKXĐ\leftrightarrow\sqrt{x}>0\leftrightarrow\sqrt{x}+2>0\)với mọi x thuộc ĐKXĐ

\(\rightarrow\)Tử thức luôn dương với mọi x thuộc ĐKXĐ

Xét mẫu thức ta có  :

 \(\sqrt{x}-2>0\) (vì \(\sqrt{x}>0\) với mọi x thuộc ĐKXĐ)

\(\leftrightarrow\sqrt{x}=2\)\(\leftrightarrow x>4\)(tm đkxđ)

Vậy..............

a: \(P=\dfrac{x+\sqrt{x}+1+11\sqrt{x}-11+34}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x+12\sqrt{x}+24}{\sqrt{x}+2}\)

b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:

\(P=\dfrac{3-2\sqrt{2}+12\left(\sqrt{2}-1\right)+24}{\sqrt{2}-1+2}\)

\(=\dfrac{27-2\sqrt{2}+12\sqrt{2}-12}{\sqrt{2}+1}=5+5\sqrt{2}\)

Bài 1 : Với : \(x>0;x\ne1\)

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

Thay vào ta được : \(P=x=25\)

Bài 2 : 

a, Với \(x\ge0;x\ne1\)

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

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

Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)

Bài 3 : \(x\ge0;x\ne1\)

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

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

b, Ta có : \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\Rightarrow4\sqrt{x}+8=5\sqrt{x}-5\)

\(\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\)(tmđk )