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

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)

\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b: P=1/4

=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)

=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)

=>\(4\sqrt{x}-8-3\sqrt{x}=0\)

=>\(\sqrt{x}=8\)

=>x=64

c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)

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

ĐKXĐ: \(x\ge4\)

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

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

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

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

b/ A = 0 \(\Rightarrow\frac{\sqrt{x}-2}{-3\sqrt{x}}=0\Rightarrow\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

Cho mình sửa lại:

Điều kiện: x > 4

nên câu b loại x = 4 nha

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

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

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

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

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

\(đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

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

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

\(=\frac{\left(x.\sqrt{x}-1\right)\left(1-x\right)}{x\left(1-x\right)}=\frac{\sqrt{x^3}-1}{x}\)

\(b,\)\(A=\frac{\sqrt{x}^3-1}{x}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x}\)

Để A > 0 \(\Rightarrow\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x}>0\)

Mà \(x>0\)và \(x+\sqrt{x}+1>0\)( do x lớn hơn 0 )

\(\Rightarrow\sqrt{x}-1>0\)

\(\Rightarrow\sqrt{x}>1\Leftrightarrow\sqrt{x}>\sqrt{1}\Leftrightarrow x>1\)