ĐKXĐ: \(x\ne0\)

\(y=\sqrt{\frac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)

\(y=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)

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

\(y=\left|\frac{x^2+3}{x}\right|+\left|x-2\right|\)

Ta có bảng xét dấu:

Với \(x< 0,y=\frac{x^2+3}{-x}+2-x=\frac{2x^2-2x+3}{-x}\)

Với \(0< x\le2,y=\frac{x^2+3}{x}+2-x=\frac{2x+3}{x}\)

Với \(x>2,y=\frac{x^2+3}{x}+x-2=\frac{2x^2-2x+3}{x}\)

- Ta thấy ngay, với cả ba trường hợp thì \(y\in Z\Leftrightarrow x\in U\left(3\right)=\left\{-3;-1;1;3\right\}\)

đk: x khác 0

A = \(\sqrt{\dfrac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)

= \(\sqrt{\dfrac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)

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

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

TH1: x \(\ge2\)

A = \(\dfrac{x^2+3}{x}+x-2\)

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

TH2: \(0< x< 2\)

A = \(\dfrac{x^2+3}{x}-x+2\)

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

TH3: x < 0

A = \(\dfrac{x^2+3}{-x}-x+2\)

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

a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)

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

\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)

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

ĐKXĐ: x>=0; \(x\notin\left\{9;4\right\}\)\(P=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

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

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

Để P là số nguyên thì \(3⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2\in\left\{1;-1;3;-3\right\}\)

=>\(\sqrt{x}\in\left\{3;1;5;-1\right\}\)

=>\(\sqrt{x}\in\left\{3;1;5\right\}\)

=>\(x\in\left\{9;1;25\right\}\)

Kết hợp ĐKXĐ, ta được; \(x\in\left\{1;25\right\}\)

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9; x\neq 4$

\(P=\frac{-3\sqrt{x}+9}{x-9}: \left[\frac{9-x}{(\sqrt{x}-2)(\sqrt{x}+3)}+\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{(\sqrt{x}-2)(\sqrt{x}+3)}-\frac{(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)}\right]\)

\(=\frac{-3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}:\frac{9-x+x-9-(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)}\)

\(=\frac{-3}{\sqrt{x}+3}:\frac{-(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{-3}{\sqrt{x}+3}:\frac{-(\sqrt{x}-2)}{\sqrt{x}+3}\\ =\frac{-3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{-(\sqrt{x}-2)}=\frac{3}{\sqrt{x}-2}\)

Với $x\in\mathbb{Z}$, để $P$ nguyên thì $\sqrt{x}-2$ là ước nguyên của 3

$\Rightarrow \sqrt{x}-2\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow \sqrt{x}\in \left\{3; 1; 5; -1\right\}$

$\Rightarrow x\in \left\{9; 1; 25\right\}$

Theo ĐKXĐ suy ra $x=1$ hoặc $x=25$

a) \(B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\left(đk:x\ge0,x\ne4\right)\)

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

c) \(C=A\left(B-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{-2}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{1;-1;2-2\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0\right\}\)

\(\Rightarrow x\in\left\{0;1;9;16\right\}\)

ĐKCĐ: \(x\ge0;x\ne9,x\ne4\)

\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ \)

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

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

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

b, \(A\inℤ\Leftrightarrow\frac{3}{\sqrt{x}-2}\inℤ\)

Nếu x không là số chính phương thì  \(\sqrt{x}\)là số vô tỉ thì \(\sqrt{x}-2\)là số vô tỉ\(\Rightarrow A=\frac{3}{\sqrt{x}-2}\)là số vô tỉ

Nếu x là số chính phương thì \(\sqrt{x}\)là số nguyên thì \(\sqrt{x}-2\inℤ\Rightarrow\sqrt{x}-2\inƯ\left(3\right)\Rightarrow\sqrt{x}-2\in\left\{\pm1;\pm3\right\}\Rightarrow\sqrt{x}\in\left\{1;3;5\right\}\)\(\Rightarrow x\in\left\{1;9;25\right\}\)

Mà theo ĐKXĐ có x khác 9 => \(x\in\left\{1,25\right\}\)