Đáp án:
a) \( - \left( {{x^2} + 2x + 2} \right)\)
b) Max=-1
Giải thích các bước giải:
\(\begin{array}{l}
a)C = \dfrac{{{{\left( {x + 2} \right)}^2}}}{x}.\dfrac{{x + 2 - {x^2}}}{{x + 2}} - \dfrac{{{x^2} + 6x + 4}}{x}\\
= \dfrac{{\left( {x + 2} \right)}}{x}.\dfrac{{x + 2 - {x^2}}}{1} - \dfrac{{{x^2} + 6x + 4}}{x}\\
= \dfrac{{{x^2} + 2x - {x^3} + 2x + 4 - 2{x^2} - {x^2} - 6x - 4}}{x}\\
= \dfrac{{ - {x^3} - 2{x^2} - 2x}}{x}\\
= - {x^2} - 2x - 2\\
= - \left( {{x^2} + 2x + 2} \right)\\
b)C = - \left( {{x^2} + 2x + 2} \right)\\
= - \left( {{x^2} + 2x + 1 + 1} \right)\\
= - {\left( {x + 1} \right)^2} - 1\\
Do:{\left( {x + 1} \right)^2} \ge 0\forall x\\
\to - {\left( {x + 1} \right)^2} \le 0\\
\to - {\left( {x + 1} \right)^2} - 1 \le - 1\\
\to Max = - 1\\
\Leftrightarrow x + 1 = 0\\
\to x = - 1
\end{array}\)
