Đáp án:
d) \(Min = - \dfrac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{{x^2} + 2x + 4 - {x^2} - 8 - 4\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{2x - 4 - 4x + 8}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{ - 2x + 4}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{ - 2\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= - \dfrac{2}{{{x^2} + 2x + 4}}\\
b)2{x^2} + x - 6 = 0\\
\to \left( {2x - 3} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - 2
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
P = - \dfrac{2}{{{{\left( {\dfrac{3}{2}} \right)}^2} + 2.\left( {\dfrac{3}{2}} \right) + 4}} = - \dfrac{8}{{37}}\\
P = - \dfrac{2}{{{{\left( { - 2} \right)}^2} + 2.\left( { - 2} \right) + 4}} = - \dfrac{1}{2}
\end{array} \right.\\
c)Do:{x^2} + 2x + 1 + 3 > 0\forall x\\
\to \dfrac{2}{{{x^2} + 2x + 4}} < 0\\
\to - \dfrac{2}{{{x^2} + 2x + 4}} > 0\\
\to P > 0\\
d)Do:{x^2} + 2x + 1 + 3 = {\left( {x + 1} \right)^2} + 3\\
Do:{\left( {x + 1} \right)^2} \ge 0\forall x\\
\to {\left( {x + 1} \right)^2} + 3 \ge 3\\
\to \dfrac{2}{{{{\left( {x + 1} \right)}^2} + 3}} \le \dfrac{2}{3}\\
\to - \dfrac{2}{{{{\left( {x + 1} \right)}^2} + 3}} \ge - \dfrac{2}{3}\\
\to Min = - \dfrac{2}{3}\\
\Leftrightarrow x = - 1
\end{array}\)
