a: ĐKXĐ: x>0; x<>1

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

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

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

b: Q>2

=>Q-2>0

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

=>căn x-1>0

=>x>1

a) ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

b) Q>2 <=> \(\dfrac{x}{\sqrt{x}-1}>2\Leftrightarrow x>2\sqrt{x}-2\)

\(\Leftrightarrow x-2\sqrt{x}+2>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+1>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le1\end{matrix}\right.\)

KL:.....

Với \(x\ge0;x\ne4\)

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

\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}-\sqrt{x}-2-3\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)

\(=\dfrac{2x-4\sqrt{x}}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)

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

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

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

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

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

\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+7}{4-x}\left(x>0;x\ne4\right)\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ P=\dfrac{\sqrt{x}+6-x-x-3\sqrt{x}-2+2\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ P=\dfrac{-2x+11}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ P=\dfrac{-2x\sqrt{x}+11\sqrt{x}+\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}\left(x-4\right)}\)

\(P=\dfrac{-2x\sqrt{x}+11\sqrt{x}+x\sqrt{x}-4\sqrt{x}+2x-8}{\sqrt{x}\left(x-4\right)}\\ P=\dfrac{-x\sqrt{x}+8\sqrt{x}+2x-8}{\sqrt{x}\left(x-4\right)}\)

\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+7}{x-4}\right)\)

\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)

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

\(=\dfrac{-x+8\sqrt{x}-15+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{-x+8\sqrt{x}-15+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(ĐK:x\ge0;x\ne4\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}-5\right)+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{8\sqrt{x}-15-x+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

Ta có: \(\dfrac{2\sqrt{x}}{\sqrt{x^3}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\)

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

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

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

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

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

Ta có:\(\dfrac{2\sqrt{x}}{\sqrt{x^3}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\)

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

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

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

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

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

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

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

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

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

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

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

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

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

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

=>\(2x-2\sqrt{x}-1=0\)

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

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

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

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

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

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

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

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

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

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

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

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

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

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

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

=>\(2x-2\sqrt{x}-1=0\)

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

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

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

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

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)