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)

1: Khi x=36 thì \(A=\dfrac{6}{2\cdot6-4}=\dfrac{6}{12-4}=\dfrac{6}{8}=\dfrac{3}{4}\)

2: 

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >4\end{matrix}\right.\)

\(C=B:A\)

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

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

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

3: \(C\cdot\sqrt{x}< \dfrac{4}{3}\)

=>\(\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{3}< 0\)

=>\(\dfrac{2\sqrt{x}\cdot3-4\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}< 0\)

=>\(6\sqrt{x}-4\sqrt{x}-8< 0\)

=>\(2\sqrt{x}-8< 0\)

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

=>\(0< =x< 16\)

Kết hợp ĐKXĐ của C, ta được: \(\left\{{}\begin{matrix}0< x< 16\\x< >4\end{matrix}\right.\)

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:

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

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

Sửa đề: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)

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

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a+3\sqrt{a}+2-a+3\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

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

\(=\dfrac{a-4}{6a\left(\sqrt{a}-1\right)}\)

c) Thay \(a=9-4\sqrt{5}\) vào Q, ta được:

\(Q=\dfrac{5-4\sqrt{5}}{6\left(9-4\sqrt{5}\right)\left(\sqrt{5}-3\right)}\)

\(=\dfrac{5-4\sqrt{5}}{6\left(9\sqrt{5}-27-20+12\sqrt{5}\right)}\)

\(=\dfrac{5-4\sqrt{5}}{6\left(21\sqrt{5}-47\right)}\)

\(=\dfrac{\left(5-4\sqrt{5}\right)\left(21\sqrt{5}+47\right)}{-24}\)

\(=\dfrac{105\sqrt{5}+235-420-188\sqrt{5}}{-24}\)

\(=\dfrac{-83\sqrt{5}-185}{-24}=\dfrac{83\sqrt{5}+185}{24}\)

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 \(\sqrt{x}+6⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1+5⋮\sqrt{x}+1\)

=>\(5⋮\sqrt{x}+1\)

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

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

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

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

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

\(=\sqrt{a}+2+\sqrt{a}+2=2\sqrt{a}+4\)

b) \(P=a+1\Rightarrow2\sqrt{a}+4=a+1\Rightarrow a-2\sqrt{a}-3=0\)

\(\Rightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-3\right)=0\) mà \(\sqrt{a}+1>0\Rightarrow\sqrt{a}=3\Rightarrow a=9\)

a) Ta có: \(P=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{2-\sqrt{a}}\)

\(=\sqrt{a}+2+2+\sqrt{a}\)

\(=2\sqrt{a}+4\)