c) Ta có: \(C=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{x-4}\)

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

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

d)

Sửa đề: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)

Ta có: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)

\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

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

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

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

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

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

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

a: \(A=3+\left(-2\right)\cdot\sqrt{3}+3\cdot\sqrt{3}-2-\sqrt{3}\)

\(=3-2=1\)

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

b: B<A

=>B-1<0

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

=>-1/căn x<0

=>căn x>0

=>x>0 và x<>1

\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))

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

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

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

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

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

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

1. ĐKXĐ:...

\(8-2x-\dfrac{2}{x}-2\sqrt{2-x^2}-2\sqrt{2-\dfrac{1}{x^2}}=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{x}-1=0\\\sqrt{2-x^2}-1=0\\\sqrt{2-\dfrac{1}{x^2}}-1=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

2.

ĐKXĐ:...

Ta có:

\(VT=x\sqrt{x}+1.\sqrt{12-x}\le\sqrt{\left(x^2+1\right)\left(x+12-x\right)}=2\sqrt{3\left(x^2+1\right)}\)

Dấu "=" xảy ra khi và chỉ khi: \(x\sqrt{12-x}=\sqrt{x}\)

\(\Leftrightarrow x^3-12x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=6-\sqrt{35}\\x=6+\sqrt{35}\end{matrix}\right.\)

3. ĐKXĐ: ...

Với \(x=0\) ko phải nghiệm

Với \(x>0\) pt tương đương:

\(\left(\dfrac{x+8\sqrt{x}+4}{\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}+4}{\sqrt{x}}\right)=36\)

\(\Leftrightarrow\left(\sqrt{x}+\dfrac{4}{\sqrt{x}}+8\right)\left(\sqrt{x}+\dfrac{4}{\sqrt{x}}-1\right)=36\)

Đặt \(\sqrt{x}+\dfrac{4}{\sqrt{x}}-1=t\ge3\)

\(t\left(t+9\right)=36\Leftrightarrow t^2+9t-36=0\)

\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-12\left(loại\right)\end{matrix}\right.\)

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

\(\Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)

\(\Leftrightarrow x=4\)

a) Ta có: \(P=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

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

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

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

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=5+\sqrt{2}-4-\sqrt{2}\)

=1

Thay x=1 vào P, ta được:

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

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

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

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

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

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

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

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

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

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

Bài 1:
1) A = \(10\sqrt{\dfrac{1}{5}}\) - \(3\sqrt{\left(2-\sqrt{5}\right)^2}\) + \(\sqrt{5}\)

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

       = \(3\sqrt{5}\) - \(3\sqrt{5}\) + 6 + \(\sqrt{5}\)

        = 6

     B = \(\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\) \(\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\) (ĐKXĐ: x > 0; x ≠ 4)

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

        = \(\dfrac{-4x-8\sqrt{x}}{x-4}\cdot\dfrac{x-2\sqrt{x}}{3-\sqrt{x}}\)

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

        = \(-\dfrac{4x}{3-\sqrt{x}}\)

        = \(\dfrac{4x}{\sqrt{x}-3}\)

Bài 2: \(\sqrt{4x^2-4x+1}=\sqrt[3]{27}\)

     ⇔ \(\sqrt{\left(2x-1\right)^2}=3\)

     ⇔ \(\left|2x-1\right|=3\)

     ⇔ \(\left\{{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)

     ⇔ \(\left\{{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)

     ⇔ \(\left\{{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Vậy S = {2; -1}

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

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

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

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49