Lời giải:
a) ĐK: \(x\neq 1\)
PT \(\Leftrightarrow \frac{x^2+x+1}{(x-1)(x^2+x+1)}+\frac{x-1}{(x-1)(x^2+x+1)}=\frac{1}{x^3-1}\)
\(\Leftrightarrow \frac{x^2+x+1}{x^3-1}+\frac{x-1}{x^3-1}=\frac{1}{x^3-1}\)
\(\Rightarrow x^2+x+1+x-1=1\)
\(\Leftrightarrow x^2+2x=1\)
\(\Leftrightarrow (x+1)^2=2\Rightarrow \left[\begin{matrix} x+1=\sqrt{2}\rightarrow x=\sqrt{2}-1\\ x+1=-\sqrt{2}\rightarrow x=-\sqrt{2}-1\end{matrix}\right.\)
b)
PT \(\Leftrightarrow 1-\frac{1}{x^2-2x+2}+1-\frac{2}{x^2-2x+3}=2-\frac{6}{x^2-2x+4}\)
\(\Leftrightarrow \frac{x^2-2x+1}{x^2-2x+2}+\frac{x^2-2x+1}{x^2-2x+3}=\frac{2x^2-4x+2}{x^2-2x+4}\)
\(\Leftrightarrow \frac{(x-1)^2}{x^2-2x+2}+\frac{(x-1)^2}{x^2-2x+3}=\frac{2(x-1)^2}{x^2-2x+4}\)
\(\Leftrightarrow (x-1)^2\left(\frac{1}{x^2-2x+2}+\frac{1}{x^2-2x+3}-\frac{2}{x^2-2x+4}\right)=0\)
Vì \(x^2-2x+4> x^2-2x+3> x^2-2x+2>0\)
\(\Rightarrow \frac{1}{x^2-2x+4}< \frac{1}{x^2-2x+3}< \frac{1}{x^2-2x+2}\)
\(\Rightarrow \frac{1}{x^2-2x+2}+\frac{1}{x^2-2x+3}-\frac{2}{x^2-2x+4}>0\)
Do đó \((x-1)^2=0\Rightarrow x=1\)
Vậy.........
