Question

Cho B=\(\frac{\sqrt{x}-5}{\sqrt{x}+1}\). Tìm x để B \(\in\) Z

Answer 1

Ta có: B = \(\frac{\sqrt{x}-5}{\sqrt{x}+1}\) = \(\frac{\sqrt{x}+1-1-5}{\sqrt{x}+1}\) = \(\frac{\sqrt{x}+1-6}{\sqrt{x}+1}\) = \(\frac{\sqrt{x}+1}{\sqrt{x}+1}+\frac{-6}{\sqrt{x}+1}\) = 1 + \(\frac{-6}{\sqrt{x}+1}\)

\(\Rightarrow\) Để B \(\in\) Z thì -6 \(⋮\) \(\sqrt{x}+1\) \(\Rightarrow\sqrt{x}+1\inƯ\left(-6\right)\)

Mà Ư(-6) = {-6; -1; 1; 6}

* \(\sqrt{x}+1\) = -6

\(\Rightarrow\) \(\sqrt{x}\) = -7

\(\Rightarrow\) x = 49

* \(\sqrt{x}+1\) = -1

\(\Rightarrow\sqrt{x}\) = -2

\(\Rightarrow\) x = 4

* \(\sqrt{x}+1\) = 1

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

\(\Rightarrow\) x = 0

* \(\sqrt{x}+1\) = 6

\(\Rightarrow\sqrt{x}\) = 5

\(\Rightarrow\) x = 25

Vậy để B = \(\frac{\sqrt{x}-5}{\sqrt{x}+1}\) \(\in\) Z thì x = {0; 4; 25; 49}

Answer 2

để B thuộc Z => \(\frac{\sqrt{x}-5}{\sqrt{x}+1}\) là số nguyên

=> \(\sqrt{x}-5⋮\sqrt{x}+1\)

=> \(\sqrt{x}-5-\left(\sqrt{x}+1\right)⋮\sqrt{x}+1\\ \Rightarrow-6⋮\sqrt{x}+1\)

=> \(\sqrt{x}+1\inƯ_{\left(-6\right)}=\left\{1;-1;6;-6\right\}\)

ta có bảng sau:

\(\sqrt{x}+1\)	1	-1	6	-6
\(\sqrt{x}\)	0	-2	5	-7
x	0	loại	25	loại

vậy x = { 0; 25 }