Question

Tìm số thực x để biểu thức có giá trị là số nguyên

a,\(\frac{\sqrt{x}+5}{\sqrt{x}+1}\)

b,\(\frac{2}{\sqrt{x}+2}\)

 

Answer 1

a) Đặt \(A=\frac{\sqrt{x}+5}{\sqrt{x}+1}\)(1)

\(\left(1\right)\Leftrightarrow\frac{\sqrt{x}+1+4}{\sqrt{x}+1}=1+\frac{4}{\sqrt{x}+1}\)

\(A\inℤ\Leftrightarrow\frac{4}{\sqrt{x}+1}\inℤ\Leftrightarrow4⋮\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Mà \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\)nên \(\sqrt{x}+1\in\left\{1;2;4\right\}\)

\(TH1:\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

\(TH2:\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

\(TH2:\sqrt{x}+1=4\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)

Vậy \(x\in\left\{0;1;9\right\}\)thì \(\frac{\sqrt{x}+5}{\sqrt{x}+1}\)đạt giá trị nguyên

Answer 2

b) \(\frac{2}{\sqrt{x}+2}\inℤ\Leftrightarrow2⋮\left(\sqrt{x}+2\right)\)

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

Vì \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\)nên \(\sqrt{x}+2=2\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

Vậy x = 0 thì \(\frac{2}{\sqrt{x}+2}\inℤ\)

Answer 3

\(a,\frac{\sqrt{x}+5}{\sqrt{x}+1}=\frac{\sqrt{x}+1+4}{\sqrt{x}+1}\) \(ĐKXĐ:x\ge0\)

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

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

Để \(1+\frac{4}{\sqrt{x}+1}\in Z\Rightarrow\sqrt{x}+1\inƯ\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow\sqrt{x}=\left\{0;1;3\right\}\)

\(\Rightarrow x=\left\{0;1;9\right\}\)

\(b,\frac{2}{\sqrt{x}+2}\in Z\Rightarrow\sqrt{x}+2\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\)

\(\Rightarrow\sqrt{x}=\left\{-1;0;-3;-4\right\}\)

\(\Rightarrow x=0\)