Question

tìm x thuộc z để các biểu thức sau nguyên

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

\(2)B=\frac{2\sqrt{x}-1}{\sqrt{x}-3}\)

\(3)C=\frac{3\sqrt{x}-4}{\sqrt{x}+2}\)

4)\(D=\frac{3-2\sqrt{x}}{\sqrt{x}-2}\)

Answer 1

c, ĐKXĐ: \(x\ge0\)

\(C=\frac{3\sqrt{x}-4}{\sqrt{x}+2}=3-\frac{10}{\sqrt{x}+2}\in Z\)

\(\Leftrightarrow\sqrt{x}+2\in U_{10}=\left\{\pm1;\pm2;\pm5\right\}\)

Mà \(\sqrt{x}+2\ge2\Rightarrow...\)

d, ĐKXĐ: \(x\ge0;x\ne4\)

\(D=\frac{3-2\sqrt{x}}{\sqrt{x}-2}=-2-\frac{1}{\sqrt{x}-2}\in Z\)

\(\Leftrightarrow\sqrt{x}-2\in U_1=\left\{\pm1\right\}\)

Mà \(\sqrt{x}-2\ge-2\Rightarrow...\)

Answer 2

a, ĐKXĐ: \(x\ge0\)

\(A=\frac{\sqrt{x}-3}{\sqrt{x}+1}=1-\frac{4}{\sqrt{x}+1}\in Z\)

\(\Leftrightarrow\sqrt{x}+1\in U_4=\left\{\pm1;\pm2;\pm4\right\}\)

Mà \(\sqrt{x}+1\ge1\Rightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\Rightarrow x\in\left\{0;1;9\right\}\)

b, ĐKXĐ: \(x\ge0;x\ne9\)

\(B=\frac{2\sqrt{x}-1}{\sqrt{x}-3}=2+\frac{5}{\sqrt{x}-3}\in Z\)

\(\Leftrightarrow\sqrt{x}-3\in U_5=\left\{\pm1;\pm5\right\}\)

Mà \(\sqrt{x}-3\ge-3\Rightarrow\sqrt{x}-3\in\left\{\pm1;5\right\}\Rightarrow x\in\left\{4;16;25\right\}\)