Đáp án:
\(\begin{array}{l}
a,\\
A = 0\\
b,\\
B = \dfrac{{x - 2}}{{{x^2} + 2x + 4}}\\
c,\\
{P_{\min }} = 1 \Leftrightarrow x = - 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DKXD:\,\,{x^3} + 8 \ne 0 \Leftrightarrow {x^3} \ne - 8 \Leftrightarrow x \ne - 2\\
a,\\
x = 2 \Rightarrow A = \dfrac{{x - 2}}{3} = \dfrac{{2 - 2}}{3} = \dfrac{0}{3} = 0\\
b,\\
B = \dfrac{{{x^2} - 4}}{{{x^3} + 8}} = \dfrac{{{x^2} - {2^2}}}{{{x^3} + {2^3}}} = \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right).\left( {{x^2} + x.2 + {2^2}} \right)}}\\
= \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {{x^2} + 2x + 4} \right)}} = \dfrac{{x - 2}}{{{x^2} + 2x + 4}}\\
c,\\
P = A:B = \dfrac{{x - 2}}{3}:\dfrac{{x - 2}}{{{x^2} + 2x + 4}}\\
= \dfrac{{x - 2}}{3}.\dfrac{{{x^2} + 2x + 4}}{{x - 2}} = \dfrac{{{x^2} + 2x + 4}}{3}\\
{x^2} + 2x + 4 = \left( {{x^2} + 2x + 1} \right) + 3 = \left( {{x^2} + 2.x.1 + {1^2}} \right) + 3 = {\left( {x + 1} \right)^2} + 3\\
{\left( {x + 1} \right)^2} \ge 0,\,\,\,\forall x \ne - 2\\
\Rightarrow {\left( {x + 1} \right)^2} + 3 \ge 3,\,\,\,\forall x \ne - 2\\
\Leftrightarrow \left( {{x^2} + 2x + 4} \right) \ge 3,\,\,\,\forall x \ne - 2\\
\Rightarrow P = \dfrac{{{x^2} + 2x + 4}}{3} \ge \dfrac{3}{3} = 1,\,\,\,\forall x \ne - 2\\
\Rightarrow {P_{\min }} = 1 \Leftrightarrow {\left( {x + 1} \right)^2} = 0 \Leftrightarrow x + 1 = 0 \Leftrightarrow x = - 1\\
Vậy\,\,\,{P_{\min }} = 1 \Leftrightarrow x = - 1
\end{array}\)
