\(A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\) ĐKXĐ : x > 0 , x khác 9 

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

\(A=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

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

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

a) ĐKXĐ : x>hoặc = 0 ; x khác 9

Còn câu b,c,d để vài bữa mình làm tiếp cho bây giờ mình đi ngủ đã buồn ngủ quá !

                        ----------------- -Học tốt-----------------

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 )

a/ 2x-x²>0

\(\Leftrightarrow\) x(2-x)>0

\(\Leftrightarrow\) 0<x<2

b/ \(\left\{{}\begin{matrix}x-3>0\\5-x>0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x>3\\x< 5\end{matrix}\right.\)\(\Leftrightarrow\) 3<x<5

c/ x²-5x+6>0

\(\Leftrightarrow\) (x-3)(x-2)>0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x>3\\x< 2\end{matrix}\right.\)

d/ \(\left\{{}\begin{matrix}6x-1>0\\x+3>0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>\frac{1}{6}\\x>-3\end{matrix}\right.\)

\(\Leftrightarrow\) x > \(\frac{1}{6}\)

\(a,\)\(\frac{1}{1-\sqrt{x^2-3}}\)

\(đkxđ\Leftrightarrow\orbr{\begin{cases}x^2-3\ge0\\x^2-3\ne1\end{cases}}\).

\(x^2-3\ne1\)\(\Rightarrow x^2\ne4\)\(\Rightarrow x\ne\pm2\)

\(x^2-3\ge0\)\(\Rightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\ge0\)

Chia trường hợp ra làm nốt nhé 

....

\(b,\)\(\frac{x-1}{2-\sqrt{3x+1}}\)

\(đkxđ\Leftrightarrow\orbr{\begin{cases}3x+1\ge0\\\sqrt{3x+1}\ne2\end{cases}}\)

\(3x+1\ge0\)\(\Rightarrow3x\ge-1\)

\(\Rightarrow x\ge\frac{-1}{3}\)

\(\sqrt{3x+1}\ne2\)\(\Rightarrow|3x+1|\ne4\)\(\Rightarrow\hept{\begin{cases}3x-1\ne4\\3x-1\ne-4\end{cases}\Rightarrow\hept{\begin{cases}3x\ne5\\3x\ne-3\end{cases}\Rightarrow}\hept{\begin{cases}x\ne\frac{5}{3}\\x\ne-1\end{cases}}}\)

\(\Rightarrow x\ge-\frac{1}{3}\)và \(x\ne\frac{5}{3}\)