Đáp án:
c) \(Max = \dfrac{8}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne 1\\
P = \left[ {\dfrac{{{x^2} + 2 + x\left( {x - 1} \right) - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}} \right].\dfrac{2}{{x - 1}}\\
= \dfrac{{1 - x + {x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{2}{{x - 1}}\\
= \dfrac{{{x^2} - 2x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{2}{{x - 1}}\\
= \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{2}{{x - 1}}\\
= \dfrac{2}{{{x^2} + x + 1}}\\
\to dpcm\\
b){x^2} - 2x - 3 = 0\\
\to \left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
P = \dfrac{2}{{{3^2} + 3 + 1}} = \dfrac{2}{{13}}\\
P = \dfrac{2}{{{{\left( { - 1} \right)}^2} - 1 + 1}} = 2
\end{array} \right.\\
c)P = \dfrac{2}{{{x^2} + x + 1}} = \dfrac{2}{{{x^2} + 2x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}}}\\
= \dfrac{2}{{{{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}}\\
Do:{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0\\
\to {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\to \dfrac{2}{{{{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} \le \dfrac{8}{3}\\
\to Max = \dfrac{8}{3}\\
\Leftrightarrow x + \dfrac{1}{2} = 0\\
\Leftrightarrow x = - \dfrac{1}{2}
\end{array}\)
