Đáp án:
b. \(\dfrac{{{x^2} + x + 1}}{{{x^2} - x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne - 1\\
A = \dfrac{x}{{x + 1}} - \dfrac{{3 - 3x}}{{{x^2} - x + 1}} + \dfrac{{x + 4}}{{{x^3} + 1}}\\
= \dfrac{{x\left( {{x^2} - x + 1} \right) - \left( {3 - 3x} \right)\left( {x + 1} \right) + x + 4}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
= \dfrac{{{x^3} - {x^2} + x - 3x - 3 + 3{x^2} + 3x + x + 4}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
= \dfrac{{{x^3} + 2{x^2} + 2x + 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
= \dfrac{{\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
= \dfrac{{{x^2} + x + 1}}{{{x^2} - x + 1}}\\
c.A = \dfrac{{{x^2} + x + 1}}{{{x^2} - x + 1}}\\
Do:\left\{ \begin{array}{l}
{x^2} + x + 1 = {x^2} + 2x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x \ne - 1\\
{x^2} - x + 1 = {x^2} - 2x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} = {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x \ne - 1
\end{array} \right.\\
\to A > 0\forall x \ne - 1
\end{array}\)
