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 )

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

\(đkxđ\Leftrightarrow\hept{\begin{cases}x^2-x+1\ge0\\x^2-x+1\ne0\end{cases}\Rightarrow x^2-x+1>0}\)

Mà \(x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)với \(\forall x\)

\(\Rightarrow\)Biểu thức luôn được xác định với mọi x 

Lời giải:

a)

ĐKXĐ: \(\left\{\begin{matrix} x^2-3\geq 0\\ 1-\sqrt{x^2-3}\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2-3\geq 0\\ x^2-3\neq 1\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \left[\begin{matrix} x\geq \sqrt{3}\\ x\leq -\sqrt{3}\end{matrix}\right.\\ x^2\neq 4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \left[\begin{matrix} x\geq \sqrt{3}\\ x\leq -\sqrt{3}\end{matrix}\right.\\ x\neq \pm 2\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x\geq \sqrt{3}, x\neq 2\\ x\leq -\sqrt{3}, x\neq -2\end{matrix}\right.\)

b)

ĐKXĐ: \(\left\{\begin{matrix} 3x+1\geq 0\\ 2-\sqrt{3x+1}\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3x+1\geq 0\\ 3x+1\neq 4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{3}\\ x\neq 1\end{matrix}\right.\)

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

\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)

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

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

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

\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

b) Với  \(x=3\)( thỏa mãn ĐKXĐ ) ta có  \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)

c) A ở đâu ???? '-' 

Bài 1 :

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

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

b) Để \(A< -1\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)

\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)

\(\Leftrightarrow2\sqrt{x}< 1\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)

\(\Leftrightarrow x< \frac{1}{4}\)

Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)

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

\(Đk:\)\(x+3\ne0\Rightarrow x\ne-3\)

Và \(\frac{2x-1}{x+3}\ge0\)

Khi \(\frac{2x-1}{x+3}=0\Rightarrow2x-1=0\)

\(\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)

Khi \(\frac{2x-1}{x+3}>0\)\(\Rightarrow\orbr{\begin{cases}2x-1>0;x+3>0\\2x-1< 0;x+3< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2};x>-3\\x< \frac{1}{2};x< -3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x< -3\end{cases}}\)

Vậy căn thức xác định khi \(x\ge\frac{1}{2};x< -3\)

bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\) 

Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)

               \(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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