a, ĐKXĐ : \(\left\{{}\begin{matrix}x^3-1\ne0\\x^2+x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x-1\ne0\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ne0\end{matrix}\right.\)
=> \(x\ne1\)
- Ta có : \(A=\left(\frac{x-1}{x^2+x+1}-\frac{x^2-3x+1}{x^3-1}-\frac{1}{x-1}\right):\frac{x^2+1}{1-x}\)
=> \(A=\left(\frac{\left(x-1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}-\frac{x^2-3x+1}{x^3-1}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x^2+1}{1-x}\)
=> \(A=\left(\frac{x^2-2x+1-x^2+3x-1-x^2-x-1}{\left(x^2+x+1\right)\left(x-1\right)}\right):\frac{x^2+1}{1-x}\)
=> \(A=\left(\frac{-x^2-1}{\left(x^2+x+1\right)\left(x-1\right)}\right)\left(\frac{1-x}{x^2+1}\right)\)
=> \(A=\frac{-\left(x^2+1\right)\left(1-x\right)}{\left(x^2+x+1\right)\left(x-1\right)\left(x^2+1\right)}\)
=> \(A=\frac{1}{x^2+x+1}\)
b, Ta có : \(\frac{1}{A}=x^2+x+1\)
- Để \(\frac{1}{A}\) là số chính phương khi :
\(x^2+x+1=y^2\)
=> \(x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=y^2\)
=> \(\left(x+\frac{1}{2}\right)^2-y^2=-\frac{3}{4}\)
=> \(\left(x+\frac{1}{2}-y\right)\left(x+\frac{1}{2}+y\right)=-\frac{3}{4}\)
Vậy không tồn tại giá trị nguyên của x để \(\frac{1}{A}\) là số chính phương .
