Bài 5:

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

Để $C$ nguyên nhỏ nhất thì $\frac{1}{\sqrt{x}-2}$ là số nguyên nhỏ nhất.

$\Rightarrow \sqrt{x}-2$ là ước nguyên âm lớn nhất

$\Rightarrow \sqrt{x}-2=-1$

$\Leftrightarrow x=1$ (thỏa mãn đkxđ)

Bài 6:

$D(\sqrt{x}+1)=x-3$

$D^2(x+2\sqrt{x}+1)=(x-3)^2$

$2D^2\sqrt{x}=(x-3)^2-D^2(x+1)$ nguyên 

Với $x$ nguyên ta suy ra $\Rightarrow D=0$ hoặc $\sqrt{x}$ nguyên 

Với $D=0\Leftrightarrow x=3$ (tm)

Với $\sqrt{x}$ nguyên:

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

$D$ nguyên khi $\sqrt{x}+1$ là ước của $2$

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

$\Leftrightarrow x=0; 1$

Vì $x\neq 1$ nên $x=0$.

Vậy $x=0; 3$

Bài 6: 

Để D nguyên thì \(x-3⋮\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}+1\in\left\{1;2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;1\right\}\)

hay \(x\in\left\{0;1\right\}\)

a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(dkxd:x\ge0;x\ne1;x\ne4\right)\)

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

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

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

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

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

b) Với \(x\ge0;x\ne1;x\ne4\):

Thay \(x=3+2\sqrt{2}\) vào \(P\), ta được:

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

\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}+2}{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}-1}\)

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

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

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

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

c) Với \(x\ge0;x\ne1;x\ne4\),

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

Để \(P\) có giá trị nguyên thì \(\dfrac{3}{\sqrt{x}-1}\) có giá trị nguyên

\(\Rightarrow 3\vdots\sqrt x-1\\\Rightarrow \sqrt x-1\in Ư(3)\)

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

\(\Rightarrow\sqrt{x}\in\left\{2;4;0;-2\right\}\) mà \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}\in\left\{2;4;0\right\}\)

\(\Rightarrow x\in\left\{4;16;0\right\}\)

Kết hợp với ĐKXĐ của \(x\), ta được:

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

Vậy: ...

\(\text{#}Toru\)

\(A=\) \(\dfrac{x+2}{x-5}\)

\(=\dfrac{\left(x-5\right)+7}{x-5}\)

\(=1+\dfrac{7}{x-5}\)

để \(\dfrac{7}{x-5}\) ∈Z thì 7⋮x-5

⇒x-5∈\(\left(^+_-1,^+_-7\right)\)

Còn lại thì bạn tự tính nha

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

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

b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

hay \(x\in\left\{0;4;9\right\}\)

Bài 1:

Để biểu thức nhận giá trị nguyên thì \(3\sqrt{x}+1⋮2\sqrt{x}-1\)

\(\Leftrightarrow6\sqrt{x}+2⋮2\sqrt{x}-1\)

\(\Leftrightarrow2\sqrt{x}-1\in\left\{1;-1;5\right\}\)

\(\Leftrightarrow2\sqrt{x}\in\left\{2;0;6\right\}\)

hay \(x\in\left\{4;0;36\right\}\)

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

b) Ta có x >0  nên \(\sqrt{x}\) >0 

 <=>  \(2\sqrt{x}\)  > 0 

 <=>  \(x+2\sqrt{x}\)  > x 

 <=> \(\dfrac{x+2\sqrt{x}}{x}\)  > \(\dfrac{x}{x}\)

 hay A > 1

c) 

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

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

b: \(Q=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}+\sqrt{2}-1+1}=\dfrac{2\sqrt{2}-1}{7}\)