Lời giải:

a.

\(A=\frac{(x\sqrt{x}-4x)-(\sqrt{x}-4)}{2(\sqrt{x}-4)(\sqrt{x}-2)(\sqrt{x}-1)}\)

ĐKXĐ: \(\left\{\begin{matrix} x\geq 0\\ \sqrt{x}-4\neq 0\\ \sqrt{x}-2\neq 0\\ \sqrt{x}-1\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\neq 16\\ x\neq 4\\ x\neq 1\end{matrix}\right.\)

\(A=\frac{x(\sqrt{x}-4)-(\sqrt{x}-4)}{2(\sqrt{x}-4)(\sqrt{2}-2)(\sqrt{x}-1)}=\frac{(x-1)(\sqrt{x}-4)}{2(\sqrt{x}-4)(\sqrt{x}-2)(\sqrt{x}-1)}\)

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

b.

Với $x$ nguyên, để $A\in\mathbb{Z}$ thì $\sqrt{x}+1\vdots 2(\sqrt{x}-2)}$

$\Rightarrow \sqrt{x}+1\vdots \sqrt{x}-2$
$\Leftrightarrow \sqrt{x}-2+3\vdots \sqrt{x}-2$

$\Leftrightarrow 3\vdots \sqrt{x}-2$

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

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

Thử lại thấy đều thỏa mãn.

a: \(A=\dfrac{x\left(\sqrt{x}-4\right)-\left(\sqrt{x}-4\right)}{2x\sqrt{x}-8x-6x+24\sqrt{x}+4\sqrt{x}-16}\)

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

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

b: Để A nguyên thì \(2\sqrt{x}+2⋮2\sqrt{x}-4\)

\(\Leftrightarrow2\sqrt{x}-4\in\left\{2;-2;6\right\}\)

hay \(x\in\left\{9;1;25\right\}\)

a: ĐKXĐ: x=0; x<>1

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

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

b: Sửa đề: P=1/M

P=1/4-x=-1/x-4

Để P nguyên thì x-4 thuộc {1;-1}

=>x thuộc {5;3}

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

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

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

a) \(M=\left(\dfrac{2x+3\sqrt{x}}{x\sqrt{x}+1}+\dfrac{1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\left(x>0\right)\)

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

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

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

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

b) Ta có: \(\sqrt{x}+4>\sqrt{x}+1\Rightarrow\dfrac{\sqrt{x}+4}{\sqrt{x}+1}>1\)

c) \(\dfrac{\sqrt{x}+4}{\sqrt{x}+1}=1+\dfrac{3}{\sqrt{x}+1}\)

Ta có: \(\left\{{}\begin{matrix}3>0\\\sqrt{x}+1>0\end{matrix}\right.\Rightarrow1+\dfrac{3}{\sqrt{x}+1}>1\Rightarrow M>1\)

Lại có: \(\sqrt{x}+1>1\left(x>0\right)\Rightarrow\dfrac{3}{\sqrt{x}+1}< 3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}< 4\Rightarrow M< 4\)

\(\Rightarrow1< M< 4\Rightarrow M\in\left\{2;3\right\}\)

\(M=2\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=2\Rightarrow\dfrac{3}{\sqrt{x}+1}=1\Rightarrow\sqrt{x}+1=3\)

\(\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(M=3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=3\Rightarrow\dfrac{3}{\sqrt{x}+1}=2\Rightarrow2\sqrt{x}+2=3\)

\(\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

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

b: \(P=M\cdot N\)

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

\(=\dfrac{3x+3\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)

Cái này mình chỉ rút gọn được P thôi, còn P nguyên thì mình xin lỗi bạn rất nhiều nha

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