\(\left(\dfrac{2}{2-\sqrt{x}}+\dfrac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right)\) (ĐK: \(x\ne4;x>0\)) 

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

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

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

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

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

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

\(=-\dfrac{3\sqrt{x}+6-x-2\sqrt{x}}{4x}\)

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

\(\left(\dfrac{\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\) (ĐK: \(x\ge0;x\ne1;x\ne\dfrac{1}{9}\)) 

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

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

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

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

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

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

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

=2

b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{x^2}\)

1: ĐKXĐ: x>1/2

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

x^2-2x+1>=0

=>x^2>=2x-1

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

Dấu = xảy ra khi x=1

(x^2-2x+1)(x^2+2x+3)>=0

=>x^4-4x+3>=0

=>x^4>=4x-3

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

=>VT>=2

Dấu = xảy ra khi x=1

2: 4x-1=x+x+2x-1

5x-2=x+2x-1+2x-1

\(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}\right)\left(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}\right)>=9\)

=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{9}{\sqrt{x}+\sqrt{x}+\sqrt{2x-1}}\)

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

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

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

Tương tự, ta cũng có: \(\dfrac{1}{\sqrt{x}}+\dfrac{2}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{5x-2}}\)

=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\sqrt{3}\left(\dfrac{1}{\sqrt{4x-1}}+\dfrac{1}{\sqrt{5x-2}}\right)\)

Dấu = xảy ra khi x=1

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

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

\(=\dfrac{-4x-8\sqrt{x}}{x-4}=-\dfrac{4\sqrt{x}}{\sqrt{x}-2}\)

A. ĐKXĐ: $x>0; x\neq 1; x\neq 4$

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

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

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

B.

ĐKXĐ: $x\geq 0, x\neq \frac{1}{4}$

\(B=\frac{2\sqrt{x}-1+2\sqrt{x}+1}{(2\sqrt{x}+1)(2\sqrt{x}-1)}.(1-4x)=\frac{4\sqrt{x}}{4x-1}(1-4x)=-4\sqrt{x}\)

ĐKXĐ: x>0; x<>9

\(A=\left(\dfrac{-\left(\sqrt{x}+3\right)}{\sqrt{x}-3}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-\dfrac{4x}{x-9}\right):\left(\dfrac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\)

\(=\dfrac{-x-6\sqrt{x}-9+x-6\sqrt{x}+9-4x}{x-9}:\dfrac{-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-4x-12\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4x\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}-2\right)}=\dfrac{4x}{\sqrt{x}-2}\)

|A|>-A

=>A>=0

=>4x>0

=>x>0 và x<>9

\(a,P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{4-x}\right):\dfrac{x+5\sqrt{x}+6}{x-4}\left(dk:x\ge0,x\ne4\right)\)

\(=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{4x}{x-4}\right).\dfrac{x-4}{x+2\sqrt{x}+3\sqrt{x}+6}\)

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

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

\(=\dfrac{4x+8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

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

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

\(b,x=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{4}}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\\ =\sqrt{5}+2-\sqrt{5}+2\\ =4\)

Khi \(x=4\Rightarrow P=\dfrac{4\sqrt{4}}{\sqrt{4}+3}=\dfrac{4.2}{2+3}=\dfrac{8}{5}\)

\(c,P=2\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+3}=2\Leftrightarrow\dfrac{4\sqrt{x}-2\left(\sqrt{x}+3\right)}{\sqrt{x}+3}=0\Leftrightarrow2\sqrt{x}-6=0\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)

Bài 3:
a) \(\sqrt{3x-2}=4\)
⇔\(\sqrt{3x-2}=\sqrt{4^2}\)
⇔\(3x-2=4^2=16\)
    \(3x=16+2=18\)
    \(x=18:3=6\)
    Vậy \(x=6\)
b)\(\sqrt{4x^2+4x+1}-11=5\)
⇔\(\sqrt{\left(2x\right)^2+2\left(2x\right)\cdot1+1^2}-11=5\)
⇔\(\sqrt{\left(2x+1\right)^2}-11=5\)
TH1:
⇔\(\left(2x+1\right)-11=5\)
    \(2x+1=5+11=16\)
    \(2x=16-1=15\)
    \(x=15:2=7,5\)
TH2:
⇔\(\left(2x+1\right)-11=-5\)
    \(2x-1=-5+11=6\)
    \(2x=6+1=7\)
    \(x=7:2=3,5\)
    Vậy \(x=\left\{7,5;3,5\right\}\) 
    (Câu này mình không chắc chắn lắm)   
    (Học sinh lớp 6 đang làm bài này)    

Bài 4:

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

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

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

b: C-6<0

=>C<6

=>\(2\sqrt{x}< 6\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)

Bài 3

a)\(\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\)

Vậy PT có nghiệm x=6

b)\(\sqrt{4x^2+4x+1}-11=5\Leftrightarrow\sqrt{\left(2x+1\right)^2}=16\Leftrightarrow2x+1=16hoặc2x+1=-16\)

+)TH1: \(2x+1=16\Leftrightarrow x=\dfrac{15}{2}\Leftrightarrow x=7,5\)

+)TH2:\(2x+1=-16\Leftrightarrow x=\dfrac{17}{2}\Leftrightarrow x=8,5\)

Bài 4

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

\(Vậy\) \(C=2\sqrt{x}\)