\(A=\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{\sqrt{x}-1}\)  \(\left(x\ge0;x\ne1\right)\)

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

và \(B=\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{2+\sqrt{2}}{\sqrt{x}+1}\)

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

\(B=\sqrt{3}+2+\frac{1}{\sqrt{3}-\sqrt{2}}+\sqrt{2}\)

\(B=\sqrt{3}+\sqrt{2}+\frac{1}{\sqrt{3}-\sqrt{2}}+2\)

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

\(B=\frac{3-2+1}{\sqrt{3}-\sqrt{2}}+2\)

\(B=\frac{2}{\sqrt{3}-\sqrt{2}}+2\)

để A = B thì \(\sqrt{x}-1\)= \(\frac{2}{\sqrt{3}-\sqrt{2}}+2\)

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

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

\(\sqrt{x}=2\sqrt{3}+2\sqrt{2}+3\)

tới bước này tui bí :(( mong các bạn giỏi khác giúp bạn :D

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

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

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

b) \(\frac{A}{B}=\frac{\sqrt{x}+4}{\sqrt{x-1}}:\frac{1}{\sqrt{x}-1}=\sqrt{x}+4\)

Để \(\frac{A}{B}\ge\frac{x}{4}+5\)

\(\Leftrightarrow\sqrt{x}+4\ge\frac{x}{4}+5\)

\(\Leftrightarrow4\sqrt{x}+16\ge x+20\)

\(\Leftrightarrow x-4\sqrt{x}+4\le0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2\le0\)

Mà \(\left(\sqrt{x}-2\right)^2\ge0;\forall x\ge0\)

\(\Rightarrow\left(\sqrt{x}-2\right)^2=0\)

\(\Leftrightarrow x=4\)

Vậy ...

\(A-B=\frac{2\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{1-\sqrt{x}}+\frac{3\sqrt{x}-1}{x-1}\)

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

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

Để \(M< 4\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-1}< 4\)

Nếu x>=1

\(\Rightarrow3\sqrt{x}\le4\sqrt{x}-4\)

\(\Leftrightarrow4\le\sqrt{x}\)

\(\Leftrightarrow x\le16\)

Nếu x<1

\(\Rightarrow3\sqrt{x}>4\sqrt{x}-4\)

\(\Leftrightarrow4>\sqrt{x}\)

\(\Rightarrow16>x\)

Ko có x thỏa mãn

a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)

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

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

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

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

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

Lời giải:

a)

\(A=\frac{\sqrt{3}-1+\sqrt{3}+1}{(\sqrt{3}+1)(\sqrt{3}-1)}+2-\sqrt{3}=\frac{2\sqrt{3}}{3-1}+2-\sqrt{3}=\sqrt{3}+2-\sqrt{3}=2\)

b)

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

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

Ta có \(\frac{A}{B}=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\right)=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\left[\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right]=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)=\sqrt{x}+4\)

Để \(\frac{A}{B}\ge\frac{x}{4}+5\) thì \(\sqrt{x}+4\ge\frac{x}{4}+5\Leftrightarrow\sqrt{x}\ge\frac{x}{4}+1\Leftrightarrow x-4\sqrt{x}+4\le0\Leftrightarrow\left(\sqrt{x}-2\right)^2\le0\)

Mà \(\left(\sqrt{x}-2\right)^2\ge0\)

Suy ra \(\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)(tm)

Vậy x=4 thì \(\frac{A}{B}\ge\frac{x}{4}+5\)

\(B=\frac{1}{\sqrt{x}-1}\) (tự rút gọn nha)

\(\frac{A}{B}\ge\frac{x}{4}+5\\ \sqrt{x}+4\ge\frac{x}{4}+5\\ \frac{x}{4}-\sqrt{x}+1\le0\\ x-4\sqrt{x}+4\le0\\ \left(\sqrt{x}-2\right)^2\le0\\ \Rightarrow\sqrt{x}-2=0\\ \Rightarrow x=4\)

Vậy để \(\frac{A}{B}\ge\frac{x}{4}+5\) thì x=4

a, x = \(\frac{4\left(\sqrt{3}+1\right)}{3-1}-\frac{4\left(\sqrt{3}-1\right)}{3-1}\)

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

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

x = 4 (TMĐK)

=> A = \(\frac{2\sqrt{4}+1}{3\sqrt{4}+1}\)

=> A = \(\frac{5}{7}\)

Vậy x = \(\frac{4}{\sqrt{3}-1}-\frac{4}{\sqrt{3}+1}\) thì A = \(\frac{5}{7}\)

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

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

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

c, \(\frac{B}{A}>2\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}:\frac{2\sqrt{x}+1}{3\sqrt{x}+1}\) > 2

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

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

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

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

mà \(\sqrt{x}+1>0\) \(\forall\) \(x\in\) ĐKXĐ

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

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

<=> \(x>1\)

Kết hợp ĐKXĐ : x \(\ge0\) ; x \(\ne\) 1

=> x > 1 thì \(\frac{B}{A}>2\)