ĐKXĐ: x>=0; x<>1

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

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

\(=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)

Khi \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\left(\dfrac{\sqrt{3}-1}{2}\right)^2\) thì:

\(B=\dfrac{2\cdot\dfrac{2-\sqrt{3}}{2}+2\cdot\dfrac{\sqrt{3}-1}{2}+1}{4\cdot\dfrac{\sqrt{3}-1}{2}}\)

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

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

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

\(=\dfrac{1}{x-1}\)

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

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

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

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

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

\(E=\left(\dfrac{x\sqrt{x}}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) (ĐK: \(x\ne1;x>0\))

\(E=\left[\dfrac{\left(\sqrt{x}\right)^3-1^3}{x-\sqrt{x}}-\dfrac{\left(\sqrt{x}\right)^3+1^3}{x+\sqrt{x}}\right]+\left[\dfrac{x}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right]\left[\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)

\(E=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

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

\(E=\dfrac{2\sqrt{x}}{\sqrt{x}}+\dfrac{2x+2}{\sqrt{x}}\)

\(E=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)

\(A=\left(\dfrac{\sqrt{x}}{x-\sqrt{x}}-\dfrac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}}{x+1}\right)\left(x>0,x\ne1\right)\)

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

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

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

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

Lời giải:
ĐKXĐ: $x>0; x\neq 1$

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

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

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

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

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

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

\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)

a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)

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

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

b) \(\dfrac{P}{A}\left(x-1\right)=0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)

Vậy \(S=\varnothing\)

21: ĐKXĐ: x>0; x<>1

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

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

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

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

22:
DKXĐ: x>0; x<>1

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

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

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

\(=\dfrac{x}{\sqrt{x}+1}\)

23: ĐKXĐ: x>0; x<>4

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

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

\(=\dfrac{-4\sqrt{x}+4}{4}=-\sqrt{x}+1\)

24: ĐKXĐ: x>=0; x<>1

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

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

25:

ĐKXĐ: x>=0; x<>1

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

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

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

27: ĐKXĐ: x>0; x<>4

\(P=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4x-8\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-1-2\sqrt{x}+1}\)

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

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

a

ĐK: \(1< x\ne10\)

Đặt \(t=\sqrt{x-1}\Rightarrow x=t^2+1;0< t\ne3\)

Khi đó:

\(P=\left(\dfrac{t}{3+t}+\dfrac{t^2+9}{\left(3-t\right)\left(3+t\right)}\right):\left(\dfrac{3t+1}{t^2-3t}-\dfrac{1}{t}\right)\\ =\left(\dfrac{t\left(3-t\right)+t^2+9}{\left(3-t\right)\left(3+t\right)}\right):\left(\dfrac{3t+1}{t\left(t-3\right)}-\dfrac{1}{t}\right)\\ =\dfrac{3t+9}{\left(3-t\right)\left(3+t\right)}:\dfrac{3t+1-t+3}{t\left(t-3\right)}=\dfrac{3\left(t+3\right)}{\left(3-t\right)\left(3+t\right)}:\dfrac{2t+4}{t\left(t-3\right)}\\ =\dfrac{3\left(t+3\right)}{\left(3-t\right)\left(3+t\right)}.\dfrac{t\left(t-3\right)}{2t+4}=\dfrac{-3t}{2t+4}=\dfrac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)

b

Ta có:

\(x=\sqrt{\left(\sqrt{2}+1\right)^2}-\left(\sqrt{5}+1\right)\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{5}\left|1-\sqrt{2}\right|\)

\(=\sqrt{2}+1-\left(\sqrt{5}+1\right)\left|1-\sqrt{2}\right|+\sqrt{5}\left|1-\sqrt{2}\right|\)

\(=\sqrt{2}+1-\sqrt{5}\left|1-\sqrt{2}\right|-\left|1-\sqrt{2}\right|+\sqrt{5}\left|1-\sqrt{2}\right|\\ =\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\)

Vậy \(P=\dfrac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=-\dfrac{1}{2}\)

a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)

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

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

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

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

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

b) Ta có: \(x=3-2\sqrt{2}\)

\(=2-2\cdot\sqrt{2}\cdot1+1\)

\(=\left(\sqrt{2}-1\right)^2\)

Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được: 

\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

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

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

\(=\dfrac{1}{\sqrt{2}-1}\)

\(=\sqrt{2}+1\)

Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)