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

\(P\in Z\Rightarrow\sqrt{x}+3=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)

Mà \(\sqrt{x}+3\ge3;\forall x\ge0\)

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

\(\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

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 \(x\in\left\{9;1;25\right\}\)

\(P=B:A\)

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

P nguyên

=>căn x+3 thuộc Ư(-3)

=>căn x+3 thuộc {1;-1;3;-3}

=>căn x+3=3

=>x=0

\(P=\dfrac{\left(x-9\right)+6}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+6}{\sqrt{x}+3}\\ P=\sqrt{x}-3+\dfrac{6}{\sqrt{x}+3}\in Z\\ \Leftrightarrow6⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;3\right\}\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x\in\left\{0;9\right\}\)

Lời giải:

ĐK: $x\geq 0; x\neq 4; x\neq 9$

a) 

\(P=\frac{2\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}-2)}+\frac{(2\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}-\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}-2)}\)

\(=\frac{2\sqrt{x}-9+(2\sqrt{x}+1)(\sqrt{x}-2)-(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{x-\sqrt{x}-2}{(\sqrt{x}-3)(\sqrt{x}-2)}\)

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

b) \(P=\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Với $x$ nguyên, để $P$ nguyên thì $\sqrt{x}-3$ phải là ước nguyên của $4$

Mà $\sqrt{x}-3\geq -3$ nên:

$\Rightarrow \sqrt{x}-3\in\left\{\pm 1;\pm 2;4\right\}$

$\Rightarrow x\in \left\{4;16;1;25;49\right\}$ (đều thỏa mãn.

\(P\in Z\Rightarrow3P\in Z\Rightarrow\dfrac{3\sqrt{x}+15}{3\sqrt{x}+1}\in Z\)

\(\Rightarrow1+\dfrac{14}{3\sqrt{x}+1}\in Z\)

\(\Rightarrow3\sqrt{x}+1=Ư\left(14\right)=\left\{1;2;7;14\right\}\) (do \(3\sqrt{x}+1\ge1\))

\(3\sqrt{x}+1=1\Rightarrow x=0\)

\(3\sqrt{x}+1=2\Rightarrow x=\dfrac{1}{9}\notin Z\) (loại)

\(3\sqrt{x}+1=7\Rightarrow x=4\)

\(3\sqrt{x}+1=14\Rightarrow x=\dfrac{169}{9}\notin Z\) (loại)

Thế \(x=\left\{0;4\right\}\) vào P đều thỏa mãn

Vậy ....

Lời giải:
\(\frac{2-3\sqrt{x}}{\sqrt{x}+3}=\frac{11-3(\sqrt{x}+3)}{\sqrt{x}+3}=\frac{11}{\sqrt{x}+3}-3\)

Để biểu thức đã cho nguyên thì $\frac{11}{\sqrt{x}+3}$ nguyên

Đặt $\frac{11}{\sqrt{x}+3}=t$ thì hiển nhiên $t>0$ do cả tử và mẫu đều dương.

Mà: $\sqrt{x}\geq 0\Rightarrow t=\frac{11}{\sqrt{x}+3}\leq \frac{11}{3}<4$

$\Rightarrow 0< t< 4$. Mà $t$ nguyên nên $t\in \left\{1; 2; 3\right\}$

$\sqrt{x}=\frac{11}{t}-3$. Để $x$ nguyên thì $t$ là ước của $11$

$\Rightarrow t=1$

$\sqrt{x}=\frac{11}{1}-3=8\Leftrightarrow x=64$

\(a,P=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\\ b,P=1\Leftrightarrow\sqrt{x}+1=2\sqrt{x}\\ \Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\\ c,P=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\in Z\\ \Leftrightarrow\sqrt{x}+1⋮2\sqrt{x}\\ \Leftrightarrow2\sqrt{x}+2⋮2\sqrt{x}\\ \Leftrightarrow2\sqrt{x}\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}=1\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=1\)

Biểu thức thiếu dấu. Bạn coi lại.

Lời giải:

a. ĐKXĐ: $x>0$

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

b. \(P=1\Leftrightarrow \frac{\sqrt{x}+1}{2\sqrt{x}}=1\Leftrightarrow \sqrt{x}+1=2\sqrt{x}\Leftrightarrow \sqrt{x}=1\Leftrightarrow x=1\) (tm)

c.

\(\frac{\sqrt{x}+1}{2\sqrt{x}}\in\mathbb{Z}\Rightarrow \frac{\sqrt{x}+1}{\sqrt{x}}\in\mathbb{Z}\)

\(\Leftrightarrow 1+\frac{1}{\sqrt{x}}\in\mathbb{Z}\Leftrightarrow \frac{1}{\sqrt{x}}\in\mathbb{Z}\)

Với $x$ nguyên thì \(\Rightarrow \sqrt{x}\) là ước của $1$

$\Rightarrow \sqrt{x}\in \left\{1\right\}$

$\Rightarrow x\in\left\{1\right\}$

Thử lại thấy thỏa mãn. Vậy $x=1$