1: ĐKXĐ: x∉{1;-1}
\(A=\left\lbrack\frac{x-1}{3x+\left(x-1\right)^2}-\frac{x^2-3x+1}{x^3-1}-\frac{1}{1-x}\right\rbrack:\frac{x^2+2x+1}{x-1}\)
\(=\left(\frac{x-1}{x^2+x+1}-\frac{x^2-3x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{\left(x+1\right)^2}{x-1}\)
\(=\frac{\left(x-1\right)^2-\left(x^2-3x+1\right)+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x-1}{\left(x+1\right)^2}\)
\(=\frac{x^2-2x+1-x^2+3x-1+x^2+x+1}{\left(x^2+x+1\right)\cdot\left(x+1\right)^2}=\frac{x^2+2x+1}{\left(x+1\right)^2\cdot\left(x^2+x+1\right)}=\frac{1}{x^2+x+1}\)
2: Để A nguyên thì 1⋮\(x^2+x+1\)
=>\(x^2+x+1\in\left\lbrace1;-1\right\rbrace\)
TH1: \(x^2+x+1=1\)
=>\(x^2+x=0\)
=>x(x+1)=0
=>x=0(nhận) hoặc x=-1(loại)
TH2: \(x^2+x+1=-1\)
=>\(x^2+x+2=0\)
=>\(x^2+x+\frac14+\frac74=0\)
=>\(\left(x+\frac12\right)^2+\frac74=0\) (vô lý)
Để \(\frac{2}{A}=2:\frac{1}{x^2+x+1}=2\left(x^2+x+1\right)\) nguyên thì \(x^2+x+1\in Z\)
Khi x=0 thì \(x^2+x+1=0^2+0+1=1\) là số nguyên
=>Nhận
Vậy: x=0
