Question

Tìm x để các biểu thức sau có giá trị nguyên:

a)A=\(\frac{2\left(x+\sqrt{x}+1\right)}{x+1}\)

b)B=\(\frac{x^2+1}{x^2-x+1}\)

c)C=\(\frac{8\sqrt{x}+3}{4x^2+1}\)

d)D=\(\frac{x^4+1}{\left(x^2+1\right)^2}\)

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

Answer 1

\(a\))\(A=\frac{2\left(x+\sqrt{x}+1\right)}{x+1}\)(đk:\(x\ge0\))

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

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

Ta có:\(x+1-2\sqrt{x}=\left(\sqrt{x}-1\right)^2\ge0\)

\(\Rightarrow2\sqrt{x}\le x+1\)

\(\Rightarrow0< \frac{2\sqrt{x}}{x+1}\le1\)

Để \(A\in Z\Rightarrow\frac{2\sqrt{x}}{x+1}\in Z\Rightarrow\frac{2\sqrt{x}}{x+1}=0;1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

c)\(C=\frac{8\sqrt{x}+3}{4x^2+1}\)(đk:x\(\ge0\))

Ta có:\(4x^2+1=4x^2+4+4+4\ge2\sqrt[4]{4x^2\cdot4^3}=8\sqrt{x}\)

\(\Rightarrow C\le\frac{8\sqrt{x}+3}{8\sqrt{x}-15}\)(đến đây chắc dễ rồi)