\(\sqrt{9x-9}+1=13\Leftrightarrow3\sqrt{x-1}=12\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\Leftrightarrow x=17\)

\(2.\text{bạn tự tìm đk}\)

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

\(A=\frac{2\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-2\right)=\sqrt{x}\left(\sqrt{x}-2\right)< 0\Leftrightarrow x-2\sqrt{x}< 0\Leftrightarrow\left(\sqrt{x}-1\right)^2< 1\Leftrightarrow-1< \sqrt{x}-1< 1\)
\(\Leftrightarrow0< x< 4\)

Câu 1:

\(\sqrt{9x-9}+1=13\)\(ĐKXĐ:x\ge1\)

\(\Leftrightarrow\sqrt{9\left(x-1\right)}=12\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

\(\Leftrightarrow\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=16\)

\(\Leftrightarrow x=17\)(tm ĐKXĐ)

Câu 2 

ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

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

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

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

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

b Để A có giá trị âm \(\Rightarrow\frac{1}{x-2\sqrt{x}}< 0\)

vì 1>0

\(\Rightarrow x-2\sqrt{x}< 0\)

\(\Leftrightarrow0< \sqrt{x}< 2\)

\(\Leftrightarrow0< x< 4\)

kết hợp ĐKXĐ: \(\Rightarrow1< x< 4\)

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

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

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

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

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

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)

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

Ai giải giúp mình bài 1 với bài 4 trước đi

1/ 

a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)

 b/  \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

    \(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)

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

c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)

                  \(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)

                   \(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)

                                                      Vậy x = 9/25 , x = 4

1) a) ĐKXĐ :  \(0\le x\ne\frac{1}{9}\)

b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)

\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)

c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)

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

\(=\frac{a-2\sqrt{a}+1}{a+1}:\frac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}=\frac{\left(\sqrt{a}-1\right)^2}{a+1}.\frac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}=\sqrt{a}-1\)

b) \(19-8\sqrt{3}=\left(\sqrt{3}-4\right)^2\Rightarrow P=\sqrt{\left(\sqrt{3}-4\right)^2}-1=4-\sqrt{3}-1=3-\sqrt{3}\)

c) P < 1 <=> \(\sqrt{a}-1< 1\Leftrightarrow a< 4\)

Kết hợp với điều kiện : \(P< 1\Leftrightarrow\hept{\begin{cases}0< a< 4\\a\ne1\end{cases}}\)