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

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

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

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

\(=\dfrac{-x+4\sqrt{x}-4}{x-2\sqrt{x}+4}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)^2+3}< 0\)

=>A<1

c: \(2\sqrt{x}>=0;x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3>0\)

=>A>=0 với mọi x thỏa mãn  ĐKXĐ

mà A<1

nên 0<=A<1

=>Để A nguyên thì A=0

=>x=0

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

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

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

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

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

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

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

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

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

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

d) Ta có: \(D=\left(\sqrt{x}+\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\dfrac{x}{\sqrt{xy}+y}+\dfrac{y}{\sqrt{xy}-x}-\dfrac{x+y}{\sqrt{xy}}\right)\)

\(=\left(\dfrac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\dfrac{\left(x+y\right)\left(x-y\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)

\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)

\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}-y\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{-\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

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

\(=-1\)

\(a,A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)

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

\(=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{x\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\)

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

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

\(=\dfrac{6\sqrt{x}}{\sqrt{x}-2}\)

\(C=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{\sqrt{x}-2}\) (\(x\ge0,x\ne4,x\ne9\))
\(C=\left[\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}\right].\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}.\left(\sqrt{x}+2\right)^2\)
\(C=\dfrac{2}{\sqrt{x}-2}\)
 

a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)

\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)

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

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

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

=-1

Bài 1: 

a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:

\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)

b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

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

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

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

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

c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)

\(\Leftrightarrow3-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)

Đoạn $x\sqrt{x}-a$ là sao vậy bạn? Có nhầm lẫn gì không?

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

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

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

tìm cả đk giúp mik vs

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

b.

\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)

\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)

c.

Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)

Ta có:

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

Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)

Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)