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, \(A=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}=\frac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)ĐK : \(x\ne1;x\ge0\)

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

b, Thay \(x=\frac{9}{4}\Rightarrow\sqrt{x}=\frac{3}{2}\)vào biểu thức A ta được 

\(\frac{\frac{3}{2}}{\frac{3}{2}-1}=\frac{\frac{3}{2}}{\frac{1}{2}}=3\)Vậy với x = 9/4 thì A = 3 

c, Ta có : \(A=\frac{9}{4}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-1}=\frac{9}{4}\Rightarrow4\sqrt{x}=9\sqrt{x}-9\)

\(\Leftrightarrow5\sqrt{x}=9\Leftrightarrow\sqrt{x}=\frac{9}{5}\Leftrightarrow x=\frac{81}{25}\)

Vậy với A = 9/4 thì x = 81/25 

\(ĐKXĐ=x\ne1;x>0\)

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

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

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

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

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

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

\(b,A=\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=\frac{\frac{3}{2}}{\frac{3}{2}-1}=\frac{3}{\frac{2}{\frac{1}{2}}}=3\)

\(c,\frac{5}{4}=\frac{\sqrt{x}}{\sqrt{x}-1}\)

\(5\sqrt{x}-5=4\sqrt{x}\)

\(\sqrt{x}=5< =>x=25\)

nhầm, nhìn phần c thành b nhé

c, Ta có :

 \(A=\frac{5}{4}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-1}=\frac{5}{4}\Rightarrow4\sqrt{x}=5\sqrt{x}-5\)

\(\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\)

Vậy với A = 5/4 thì x = 25 

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 : \(P< 1\Leftrightarrow\hept{\begin{cases}0< a< 4\\a\ne1\end{cases}}\)

a) ĐKXĐ: \(x\ge0\); \(1-4x\ne\)0; \(2\sqrt{x}-1\ne0\); \(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\ne\)0

<=> \(x\ge0\); x \(\ne\)1/4

Ta có:  \(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)

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

\(A=\frac{\sqrt{x}-1}{1-4x}\cdot\frac{1-4x}{6x+4x+2\sqrt{x}}\)

\(A=\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\)

b)Với x \(\ge\)0 và x \(\ne\)1/4

Ta có: A > A² <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\left(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)^2\)

<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\left(1-\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)>0\)

<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10x+2\sqrt{x}-\sqrt{x}+1}{10x+2\sqrt{x}}>0\)

<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10+\sqrt{x}+1}{10x+2\sqrt{x}}>0\)

<=> \(\sqrt{x}-1>0\) <=> \(x>1\)

c) Với x\(\ge\)0 và x \(\ne\)1/4 (1)

Ta có: \(\left|A\right|>\frac{1}{4}\) <=> \(\orbr{\begin{cases}A>\frac{1}{4}\\A< -\frac{1}{4}\end{cases}}\)

TH1: \(A>\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\frac{1}{4}\)

<=> \(4\left(\sqrt{x}-1\right)>10x+2\sqrt{x}\)

<=> \(4\sqrt{x}-4>10x+2\sqrt{x}\)

<=> \(10x-2\sqrt{x}+4< 0\)(vô liia  vì \(10x-2\sqrt{x}+4>0\))

TH2: \(A< -\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}< -\frac{1}{4}\)

<=> \(4\left(\sqrt{x}-1\right)< -10x-2\sqrt{x}\)

<=> \(4\sqrt{x}-4+10x+2\sqrt{x}< 0\)

<=> \(10x+6\sqrt{x}-4< 0\)

<=> \(5x+3\sqrt{x}-2< 0\)

<=> \(\left(5\sqrt{x}-2\right)\left(\sqrt{x}+1\right)< 0\)

<=> \(x< \frac{4}{25}\) (2)

Từ (1) và (2) => \(0\le x< \frac{4}{25}\)

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

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

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

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

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

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

b)Khi \(x=\frac{9}{4}\)

\(\Rightarrow\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=3\)

c)\(A=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)}< 1\)

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

=>ko có giá trị nào