$\begin{array}{l} P = \left( {x - \dfrac{1}{{1 - x}}} \right):\dfrac{{{x^2} - x + 1}}{{{x^2} - 2x + 1}}\\ P = \left( {\dfrac{{x\left( {1 - x} \right) - 1}}{{1 - x}}} \right).\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} - x + 1}} = \dfrac{{ - {x^2} + x - 1}}{{1 - x}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} - x + 1}}\\ P = \dfrac{{{x^2} - x + 1}}{{x - 1}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} - x + 1}} = x - 1\\ +P= x - 1 = {x^2} \Leftrightarrow {x^2} - x + 1 = 0 \Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} = 0 \end{array}$
Mà $x^2-x+1=(x-\dfrac{1}{2})^2+\dfrac{3}{4}>\dfrac{3}{4}>0$ nên không có $x$ thỏa mãn sao cho $P=x^2$
