\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\left(x\ge0;x\ne9\right)=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)

Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+3}\in Z\)

\(\Leftrightarrow2⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-5;-4;-2;-1\right\}\\ \Leftrightarrow x\in\left\{1;4;16;25\right\}\)

Vậy \(x\in\left\{1;4;16;25\right\}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\)

Tick plz

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+3\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne-3\left(loại\right)\end{matrix}\right.\)\(\Rightarrow x\ge0\)

\(x\in Z\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+1\right)⋮\left(\sqrt{x}+3\right)\)

\(\Rightarrow\left(\sqrt{x}+3-2\right)⋮\left(\sqrt{x}+3\right)\)

Vì \(\Rightarrow\left(\sqrt{x}+3\right)⋮\left(\sqrt{x}+3\right)\)

\(\Rightarrow2⋮\left(\sqrt{x}+3\right)\Rightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Ta có bảng:

Vậy không có x nguyên thỏa mãn đề bài

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

Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\) thì \(2⋮\sqrt{x}+3\Rightarrow\sqrt{x}+3\in\) Ư(2)\(=\left\{1;-1;2;-2\right\}\)

Vì \(\sqrt{x}\ge0\Rightarrow x\in\varnothing\)

Biểu thức gì vậy bạn?

tìm các giá trị nguyên của x để biểu thức P=A.B  nhận giá trị nguyên

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 \(x\in\left\{0;4;9\right\}\)

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

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

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

a = 9;0

Để A là số nguyên dương thì \(\left\{{}\begin{matrix}3\sqrt{x}+6-7⋮\sqrt{x}+2\\x>\dfrac{1}{9}\end{matrix}\right.\Leftrightarrow\sqrt{x}+2=7\)

hay x=25