Question

[Dương Tịch]

Cho A= căn x / căn x-2

Tìm x nguyên để A nhận gt nguyên

Answer 1

[I don't need you]

ta có: \(A=\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\frac{2}{\sqrt{x}-2}.\)

để A nguyên 

\(\Rightarrow\frac{2}{\sqrt{x}-2}\in Z\Rightarrow2⋮\sqrt{x}-2\Rightarrow\sqrt{x}-2\inƯ_{\left(2\right)}=[\pm1;\pm2]\) 

nếu \(\sqrt{x}-2=1\Rightarrow\sqrt{x}=3\Rightarrow x=9\left(TM\right)\)

...

bn tự xét tiếp nha!

Answer 2

[ctk_new]

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

\(A\inℤ\Leftrightarrow\frac{2}{\sqrt{x}-2}\in Z\Leftrightarrow2⋮\left(\sqrt{x}-2\right)\)

\(\Leftrightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Lập bảng:

\(\sqrt{x}-3\)	\(1\)	\(-1\)	\(2\)	\(-2\)
\(\sqrt{x}\)	\(4\)	\(2\)	\(5\)	\(1\)
\(x\)	\(16\)	\(4\)	\(25\)	\(1\)

Vậy \(x\in\left\{1;4;16;25\right\}\)