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

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

b: A=3

=>căn x-1=3

=>căn x=4

=>x=16

c: A<=5

=>căn x-1<=5

=>căn x<=6

=>0<=x<=36

=>\(x\in\left\{0;2;3;4;...;36\right\}\)

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

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

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

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

b: Để \(A>\dfrac{1}{3}\) thì \(A-\dfrac{1}{3}>0\)

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

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

hay x<9

Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)

a) \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\left(đk:x\ge0,x\ne0\right)\)

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

b) \(A>\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\)

\(\Leftrightarrow6>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\) 

a: \(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{x-4}\)

\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}\)

\(=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

b: Để A=2 thì \(3\sqrt{x}=2\sqrt{x}+4\)

hay x=16

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

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

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

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

b. Đặt \(B=A-2x\)

\(B=\sqrt{x}-1-2x=-2\left(\sqrt{x}-\dfrac{1}{4}\right)^2-\dfrac{7}{8}\le-\dfrac{7}{8}\)

\(B_{max}=-\dfrac{7}{8}\) khi \(\sqrt{x}-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)

Lời giải:

a. 

$A=\frac{\sqrt{x}(5-\sqrt{x})-(\sqrt{x}+5)(\sqrt{x}+1)}{(\sqrt{x}+5)(5-\sqrt{x})}-\frac{5-9\sqrt{x}}{(\sqrt{x}-5)(\sqrt{x}+5)}$

$=\frac{-2x-10\sqrt{x}}{(\sqrt{x}+5)(5-\sqrt{x})}$

$=\frac{-2\sqrt{x}(\sqrt{x}+5)}{(\sqrt{x}+5)(5-\sqrt{x})}=\frac{2\sqrt{x}}{\sqrt{x}-5}$

b.

$A< 1\Leftrightarrow \frac{2\sqrt{x}}{\sqrt{x}-5}<1$

$\Leftrightarrow \frac{\sqrt{x}+5}{\sqrt{x}-5}<0$

$\Leftrightarrow \sqrt{x}-5<0$

$\Leftrightarrow 0\leq x< 25$

Kết hợp với đkxđ suy ra $0\leq x< 25$

Bạn xem tại đây:

https://hoc24.vn/cau-hoi/adfracsqrtxsqrtx5-dfracsqrtx15-sqrtx-dfrac5-9sqrtxx-25-voi-xge0xne25rut-gon-a2-tim-tat-ca-cac-gia-tri-cua-x-de-a1.7900547231312

\(A=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=1+\dfrac{5}{\sqrt{x}+1}\)

Để A nguyên

⇒ \(5⋮\left(\sqrt{x}+1\right)\)

\(\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Còn lại bạn tự xét các trường hợp nha

\(\Leftrightarrow A\left(\sqrt{x}+1\right)=\sqrt{x}+6\Leftrightarrow A\sqrt{x}+A=\sqrt{x}+6\)

\(\Leftrightarrow A\sqrt{x}-\sqrt{x}=6-A\Leftrightarrow\sqrt{x}\left(A-1\right)=6-A\Leftrightarrow\sqrt{x}=\dfrac{6-A}{A-1}\)

Vì \(\sqrt{x}\ge0\Rightarrow\dfrac{6-A}{A-1}\ge0\)

TH1 : \(\left\{{}\begin{matrix}6-A\ge0\\A-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A\le6\\A\ge1\end{matrix}\right.\Leftrightarrow1\le A\le6\)

TH2 : \(\left\{{}\begin{matrix}6-A\le0\\A-1\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A\ge6\\A\le1\end{matrix}\right.\)( loại ) 

Với A = 1 => \(\sqrt{x}+1=\sqrt{x}+6\Leftrightarrow1=6\)(vô lí)

Với A = 2 => x = 16 

Với A = 3 => x = 2,25 

Với A = 4 => x \(\approx\)0,444

Với A = 5 => x = 0,0625

Với A = 6 => x= 0 

\(A=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1+5}{\sqrt{x}+1}=1+\dfrac{5}{\sqrt{x}+1}\)

Vì 1 là số nguyên nên để A nguyên thì \(\dfrac{5}{\sqrt{x}+1}\)là số nguyên

Để \(\dfrac{5}{\sqrt{x}+1}\)nguyên thì \(5⋮\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(5\right)\)

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

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

Mà \(\sqrt{x}\ge0\)\(\Rightarrow\sqrt{x}\in\left\{0;4\right\}\)

\(\Leftrightarrow x\in\left\{0;2\right\}\)(TM:x\(\ge0\))

Vậy.....