$P = ( 1 - \dfrac{{x}^{2}}{{x}^{2} - x + 1}) : \dfrac{{x}^{2} + 2x + 1}{{x}^{3} +1}$
$P = ( 1 - \dfrac{{x}^{2}}{{x}^{2} - x + 1}) : \dfrac{{x}^{2} + 2x + 1}{(x + 1)({x}^{2} - x +1)}$
a) ĐKXĐ : x + 1 $\neq$ 0 ⇔ x $\neq$ -1
b) $P = ( 1 - \dfrac{{x}^{2}}{{x}^{2} - x + 1}) : \dfrac{{x}^{2} + 2x + 1}{(x + 1)({x}^{2} - x +1)}$
⇔ $P = ( \dfrac{{x}^{2} - x + 1}{{x}^{2} - x + 1} +\dfrac{{x}^{2}}{{x}^{2} - x + 1}) - \dfrac{{x}^{2}}{{x}^{2} - x + 1}) . \dfrac{(x + 1)({x}^{2} - x +1)}{{x}^{2} + 2x + 1}$
⇔$P = ( \dfrac{{x}^{2} - x + 1 - {x}^{2}}{{x}^{2} - x + 1} . \dfrac{(x + 1)({x}^{2} - x +1)}{{x + 1}^{2}}$
⇔$P = ( \dfrac{(1 - x)(x + 1)({x}^{2} - x +1}{({x}^{2} - x + 1)({x + 1}^{2})}$
⇔$P = \dfrac{1 - x}{x + 1}$
c) P = 2 ⇔ $\dfrac{1 - x}{x + 1}$ = 2
⇔ $\dfrac{1 - x}{x + 1} - \dfrac{2(x + 1)}{x + 1}$ = 0
⇔ $1 - x - 2x - 2 = 0$
⇔ $-3x - 1 = 0$
⇔ $-3x = 1$
⇔ $ x = \dfrac{-1}{3}$
Vậy để P = 2 thì x = $\dfrac{-1}{3}$
