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