Question

[RIBFUBUG]

Tim cac so nguyen x de biet thuc \(A=\dfrac{x^5+1}{x^3+1}\) co gt la so nguyen

Answer 1

[Nguyễn Việt Lâm]

Với các giá trị nguyên của \(x\ne-1\), để A nguyên thì \(\left(x^5+1\right)⋮\left(x^3+1\right)\)

\(\Leftrightarrow\left(x^5+x^2-\left(x^2-1\right)\right)⋮\left(x^3+1\right)\)

\(\Leftrightarrow\left(x^2\left(x^3+1\right)-\left(x^2-1\right)\right)⋮\left(x^3+1\right)\)

\(\Leftrightarrow\left(x^2-1\right)⋮\left(x^3+1\right)\)

\(\Leftrightarrow\left(x-1\right)⋮\left(x^2-x+1\right)\)

\(\Rightarrow x\left(x-1\right)⋮\left(x^2-x+1\right)\)

\(\Leftrightarrow\left(x^2-x+1-1\right)⋮\left(x^2-x+1\right)\)

\(\Leftrightarrow1⋮\left(x^2-x+1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x+1=1\\x^2-x+1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\left(vn\right)\end{matrix}\right.\)

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