\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)

\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)

\(a,P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right)\left(dkxd:x\ge0,x\ne1\right)\)

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

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

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

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

\(b,x=4+2\sqrt{3}\Rightarrow P=\dfrac{\left(4+2\sqrt{3}\right)-2}{\sqrt{4+2\sqrt{3}}}\)

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

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

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

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

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

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

b: Khi x=4+2căn 3 thì \(P=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=2\)

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

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

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

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

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

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)\)

\(=3\sqrt{x}-6\)

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

\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)   (1)

ĐKXĐ: \(x>0\)

\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)

\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)

\(\Leftrightarrow3x-10\sqrt{x}+1=0\)   (2)

Đặt \(t=\sqrt{x}\ge0\)

\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)

\(\Delta'=25-4=22\)

Phương trình có hai nghiệm phân biệt:

\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)

\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)

Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)

Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)

Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)

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

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

b: P=(4căn x-1)/căn x

=>3x-6căn x-4căn x+1=0

=>3x-10căn x+1=0

=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9

a: \(P=\dfrac{x+\sqrt{x}}{x-\sqrt{x}}\cdot\dfrac{3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}-1}\)

b: Để P=1 thì \(\sqrt{x}-1=3\)

hay x=16

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

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

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

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

\(P=1\)

\(\Leftrightarrow1=\dfrac{3}{\sqrt{x}-1}\)

\(\Leftrightarrow\sqrt{x}-1=3\)

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

\(\Leftrightarrow x=16\left(tm\right)\)

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

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

b: Khi x=9 thì P=9-3+1=7

c: P=3

=>x-căn x-2=0

=>(căn x-2)(căn x+1)=0

=>x=4

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

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

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

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

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

\(b,P>\dfrac{1}{2}\Rightarrow\dfrac{x-1}{x}>\dfrac{1}{2}\Leftrightarrow2x-2>x\Leftrightarrow x>2\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

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

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

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

b: P>-1

=>P+1>0

=>\(\dfrac{4}{\sqrt{x}-3}+1>0\)

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

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}>0\)

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

=>x>9

\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\text{x > 0, x ≠ 1}\)

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

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

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

a, ĐK: \(x\ge0;x\ne1\)

\(P=\left(1+\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}-1}\right).\left(1-\dfrac{6}{\sqrt{x}+5}\right)\)

\(=\left[\dfrac{x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right].\dfrac{\sqrt{x}+5-6}{\sqrt{x}+5}\)

\(=\dfrac{x+5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}+5}\)

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

b, \(P=\dfrac{\sqrt{x}}{\sqrt{x}+1}=1-\dfrac{1}{\sqrt{x}+1}\in Z\)

\(\Leftrightarrow\sqrt{x}+1\in\left\{-1;1\right\}\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy ta có điều phải chứng minh.

a: Ta có: \(P=\left(1+\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}-1}\right)\cdot\left(1-\dfrac{6}{\sqrt{x}+5}\right)\)

\(=\dfrac{x-1+2\sqrt{x}-2+3\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+5-6}{\sqrt{x}+5}\)

\(=\dfrac{x+5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+5}\)

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