Đáp án:
$\begin{array}{l}
1c)\frac{{3x - 2}}{{2x - 1}} + \frac{{1 - x}}{{2x - 1}}\\
= \frac{{3x - 2 + 1 - x}}{{2x - 1}}\\
= \frac{{2x - 1}}{{2x - 1}}\\
= 1\\
3)a)x \ne - 1\\
P = \frac{{2 - x}}{{{x^3} + 1}} + \frac{{x - 1}}{{x + 1}}:\frac{{{x^2} - x + 1}}{{x + 1}}\\
= \frac{{2 - x}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} + \frac{{x - 1}}{{x + 1}}.\frac{{x + 1}}{{{x^2} - x + 1}}\\
= \frac{{2 - x}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} + \frac{{x - 1}}{{{x^2} - x + 1}}\\
= \frac{{2 - x + \left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
= \frac{{2 - x + {x^2} - 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
= \frac{{{x^2} - x + 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
= \frac{1}{{x + 1}}\\
b)x \ne - 1\\
P = \frac{1}{P}\\
\Rightarrow \frac{1}{{x + 1}} = x + 1\\
\Rightarrow {\left( {x + 1} \right)^2} = 1\\
\Rightarrow \left[ \begin{array}{l}
x + 1 = 1\\
x + 1 = - 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 0\left( {tmdk} \right)\\
x = - 2\left( {tmdk} \right)
\end{array} \right.
\end{array}$
Vậy x=0 hoặc x=-2 thì P=1/P
