Lời giải:

ĐKXĐ: $x>0; x\neq 1; x\neq 9$

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

\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{x-1-(x-9)}{(\sqrt{x}-3)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{8}{(\sqrt{x}-1)(\sqrt{x}-3)}\)

\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)(\sqrt{x}-3)}{8}=\frac{\sqrt{x}-3}{8\sqrt{x}}\)

Để $A<0\Leftrightarrow \frac{\sqrt{x}-3}{8\sqrt{x}}<0$

$\Leftrightarrow \sqrt{x}-3<0$ (do $8\sqrt{x}>0$)

$\Leftrightarrow \sqrt{x}<3$

$\Leftrightarrow 0\leq x< 9$

Kết hợp với đkxđ suy ra $0< x< 9; x\neq 1$

Khi $x=3-2\sqrt{2}=(\sqrt{2}-1)^2$

$\Rightarrow \sqrt{x}=\sqrt{2}-1$

Khi đó: $A=\frac{\sqrt{x}-3}{8\sqrt{x}}=\frac{\sqrt{2}-4}{8(\sqrt{2}-1)}=\frac{-2-3\sqrt{2}}{8}$

a: \(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-2x}{x-9}=\dfrac{x-3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

b: \(P=A\cdot B=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)

Để |P|>P thì P<0

=>căn x-2<0

=>0<x<4

=>x=1

\(a,ĐK:x\ne3;x\ge1\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\\ b,A=4\left(2-\sqrt{3}\right)\\ \Leftrightarrow\sqrt{x-1}+\sqrt{2}=8-4\sqrt{3}\\ \Leftrightarrow\sqrt{x-1}=8-4\sqrt{3}-\sqrt{2}\\ \Leftrightarrow x-1=\left(8-4\sqrt{3}-\sqrt{2}\right)^2\\ \Leftrightarrow x=\left(8-4\sqrt{3}-\sqrt{2}\right)^2+1=...\\ d,A=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow x-1=0\Leftrightarrow x=1\)

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

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

\(=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

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

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

b) \(A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0\)

\(\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0\)

Ta có: \(\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0\)

\(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9\)

Đề sai rồi bạn

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

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

 Để A=1/2 thì 

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

nhân chéo ta đc pt \(x-4\sqrt{x}+3=0\)

giải pt ta đc x=1 (loại)  hoặc x= 9

vậy x=9 TM

Để A<1 thì \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\Leftrightarrow2\sqrt{x}-1< \sqrt{x}+1\Leftrightarrow\sqrt{x}< 2\)

                                                                                               =>  x<4   

vậy vs 0\(\le x< 4\) và x khác 1 TM

Mình nghĩ thế này ạ

a) Với \(x\ge0,x\ne1\)ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1x}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)

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

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

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

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

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

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

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

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

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

Kết luận :

b) Với \(x\ge0,x\ne1\)ta có: \(A=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)

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

\(\Leftrightarrow\frac{4\sqrt{x}-2}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}=0\)

\(\Leftrightarrow\frac{4\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}=0\)

\(\Leftrightarrow\frac{3\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}=0\)

\(\Leftrightarrow3\sqrt{x}-3=0\)

\(\Leftrightarrow3\sqrt{x}=3\)

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

\(\Leftrightarrow x=1\)( không tm đkxđ)

Vậy không có gtri nào của x để A = \(\frac{1}{2}\)