Đáp án:
2) b) A=2
Giải thích các bước giải:
\(\begin{array}{l}
1)1 + \dfrac{{1 + 2x}}{3} \le \dfrac{{x - 1}}{6} - 2\\
\to \dfrac{{6 + 2\left( {1 + 2x} \right) - x + 1 - 2.6}}{6} \le 0\\
\to \dfrac{{3x - 3}}{6} \le 0\\
\to 3x - 3 \le 0\\
\to x \le 1\\
2)DK:x \ne \pm 1\\
A = \dfrac{{{x^2}}}{{{x^2} - 1}} - \dfrac{{2x + 1}}{{1 - {x^3}}} - \dfrac{{{x^2} + 1}}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}\\
= \dfrac{{{x^2}}}{{{x^2} - 1}} + \dfrac{{2x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{1}{{x - 1}}\\
= \dfrac{{{x^2}\left( {{x^2} + x + 1} \right) + \left( {2x + 1} \right)\left( {x + 1} \right) - \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{{x^4} + {x^3} + {x^2} + 2{x^2} + 3x + 1 - {x^3} - {x^2} - x - {x^2} - x - 1}}{{\left( {{x^2} - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{{x^4} + {x^2} + x}}{{\left( {{x^2} - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
b){x^2} + 3x + 2 = 0\\
\to \left( {x + 1} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 1\left( l \right)\\
x = - 2
\end{array} \right.\\
Thay:x = - 2\\
\to A = \dfrac{{{{\left( { - 2} \right)}^4} + {{\left( { - 2} \right)}^2} - 2}}{{\left( {{{\left( { - 2} \right)}^2} - 1} \right)\left( {{{\left( { - 2} \right)}^2} - 2 + 1} \right)}} = 2
\end{array}\)
