Đáp án:
$a.P=\dfrac{x^2}{x-1}$
$b.P\ge 4$
Giải thích các bước giải:
$\begin{split}P&=\dfrac{x^2+x}{x^2-2x+1}:(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x})\\&=\dfrac{x(x+1)}{(x-1)^2}:(\dfrac{(x-1)(x+1)}{x(x-1)}+\dfrac{x}{x(x-1)}+\dfrac{2-x^2}{x(x-1)})\\&=\dfrac{x(x+1)}{(x-1)^2}:\dfrac{(x-1)(x+1)+x+2-x^2}{x(x-1)}\\&=\dfrac{x(x+1)}{(x-1)^2}:\dfrac{(x-1)(x+1)-(x^2-2x+x-2)}{x(x-1)}\\&=\dfrac{x(x+1)}{(x-1)^2}:\dfrac{(x-1)(x+1)-(x+1)(x-2)}{x(x-1)}\\&=\dfrac{x(x+1)}{(x-1)^2}:\dfrac{x+1}{x(x-1)}\\&=\dfrac{x(x+1)}{(x-1)^2}.\dfrac{x(x-1)}{x(x+1)}\\&=\dfrac{x^2}{x-1}\end{split}$
$\begin{split}b.P&=\dfrac{x^2}{x-1}\\&=\dfrac{x^2-1+1}{x-1}\\&=\dfrac{(x-1)(x+1)+1}{x-1}\\&=x+1+\dfrac{1}{x-1}\\&=(x-1)+\dfrac{1}{x-1}+2\\&\ge 2\sqrt{(x-1).\dfrac{1}{x-1}}+2\\&=4\end{split}$
