Question

[Pox mobile]

Cho M= \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
Tìm x nguyên để M nguyên

Answer 1

[Kiyotaka Ayanokoji]

Trả lời:

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

Để \(M\inℤ\Leftrightarrow1+\frac{4}{\sqrt{x}-3}\inℤ\)

\(\Rightarrow\frac{4}{\sqrt{x}-3}\inℤ\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\sqrt{x}-3\)	\(-4\)	\(-2\)	\(-1\)	\(1\)	\(2\)	\(4\)
\(\sqrt{x}\)	\(-1\left(L\right)\)	\(1\)	\(2\)	\(4\)	\(5\)	\(7\)
\(x\)	\(/\)	\(1\left(TM\right)\)	\(4\left(TM\right)\)	\(16\left(TM\right)\)	\(25\left(TM\right)\)	\(49\left(TM\right)\)

Vậy \(x\in\left\{1,4,16,25,49\right\}\) thì \(M\inℤ\)

Answer 2

[Edogawa Conan]

Đk: x \(\ge\)0; x \(\ne\)9

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

Để M nguyên <=> \(\frac{4}{\sqrt{x}-3}\in Z\)

<=> \(4⋮\sqrt{x}-3\)<=> \(\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Do \(\sqrt{x}-3\ge-3\) => \(\sqrt{x}-3\in\left\{\pm1;\pm2;4\right\}\)

Lập bảng: 

\(\sqrt{x}-3\)	1	-1	2	-2	4
x	16	4	25	1	49

Vậy ....