a: \(B=\dfrac{x-2\sqrt{x}}{\sqrt{x}-2}-\dfrac{2x+12\sqrt{x}+18}{\sqrt{x}+3}\)

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

=căn x-2(căn x+3)

=-căn x-6

b: B+8>0

=>-căn x-6+8>0

=>-căn x+2>0

=>-căn x>-2

=>căn x<2

=>0<=x<4

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right)\cdot\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-3+2\sqrt{x}}{x-9}\cdot\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)

\(=\dfrac{3\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\cdot\dfrac{2}{\sqrt{x}-3}=\dfrac{6}{\sqrt{x}-3}\)

a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=5-1=4\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

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

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

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

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

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

c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)

d: |B|=A

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

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

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

=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)

\(M=\left(\dfrac{3}{\sqrt{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt{x}-5}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

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

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

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

a: Khi x=16 thì \(A=\dfrac{6}{16-3\cdot4}=\dfrac{6}{4}=\dfrac{3}{2}\)

b: P=A:B

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

\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{6}\)

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

c: \(P-1=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}}=\dfrac{3}{\sqrt{x}}>0\)

=>P>1

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

ĐKXĐ: x>=0

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

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

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

Thay x=4 vào A, ta được:

\(A=\dfrac{4-2+1}{2+1}=\dfrac{5-2}{3}=1\)

b: M=A*B

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\left(\dfrac{2x+6\sqrt{x}+7}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\left(\dfrac{2x+6\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right)\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{2x+6\sqrt{x}+7-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)^2}=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)

Để M>2 thì M-2>0

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

=>\(-\sqrt{x}+4>0\)

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

=>\(\sqrt{x}< 4\)

=>0<=x<16

c: Để M là số nguyên thì \(\sqrt{x}+6⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1+5⋮\sqrt{x}+1\)

=>\(5⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\{1;-1;5;-5\right\}\)

=>\(\sqrt{x}\in\left\{0;-2;4;-6\right\}\)

=>\(\sqrt{x}\in\left\{0;4\right\}\)

=>\(x\in\left\{0;16\right\}\)

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

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

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

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

2: Thay x=9 vào A, ta được:

\(A=\dfrac{3}{3+1}=\dfrac{3}{4}\)

1: =>x^2-x=3-x

=>x^2=3

=>x=căn 3 hoặc x=-căn 3

2: =>x^2-4x+3=x^2-4x+4 và x>=2

=>3=4(vô lý)

3: =>2|x-1|=6

=>|x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=-2 hoặc x=4

4: =>|2x-3|=|x-2|

=>2x-3=x-2 hoặc 2x-3=-x+2

=>x=1 hoặc x=5/3

5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)

=>x+2=0

=>x=-2