Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = 2{x^2} - 2\\
{x^2} \ge 0,\,\,\,\forall x \Rightarrow 2{x^2} \ge 0,\,\,\forall x\\
\Rightarrow 2{x^2} - 2 \ge - 2,\,\,\,\forall x\\
\Rightarrow A \ge - 2,\,\,\,\forall x\\
\Rightarrow {A_{\min }} = - 2 \Leftrightarrow {x^2} = 0 \Leftrightarrow x = 0\\
b,\\
B = \dfrac{{\left| x \right| + 2017}}{{2018}}\\
\left| x \right| \ge 0,\,\,\forall x \Rightarrow \left| x \right| + 2017 \ge 2017,\,\,\forall x\\
\Rightarrow \dfrac{{\left| x \right| + 2017}}{{2018}} \ge \dfrac{{2017}}{{2018}},\,\,\,\forall x\\
\Rightarrow B \ge \dfrac{{2017}}{{2018}},\,\,\,\forall x\\
c,\\
C = 3 - {\left( {x + 1} \right)^2}\\
{\left( {x + 1} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow 3 - {\left( {x + 1} \right)^2} \le 3,\,\,\,\forall x\\
\Rightarrow C \le 3,\,\,\,\forall x\\
\Rightarrow {C_{\max }} = 3 \Leftrightarrow {\left( {x + 1} \right)^2} = 0 \Leftrightarrow x = - 1\\
d,\\
D = - \left| {0,1 + x} \right| - 1,9 = - \left( {\left| {0,1 + x} \right| + 1,9} \right)\\
\left| {0,1 + x} \right| \ge 0,\,\,\,\forall x\\
\Rightarrow \left| {0,1 + x} \right| + 1,9 \ge 1,9,\,\,\,\forall x\\
\Rightarrow - \left( {\left| {0,1 + x} \right| + 1,9} \right) \le - 1,9,\,\,\,\,\forall x\\
\Rightarrow D \le - 1,9,\,\,\,\forall x\\
\Rightarrow {D_{\max }} = - 1,9 \Leftrightarrow \left| {0,1 + x} \right| = 0 \Leftrightarrow x = - 0,1\\
e,\\
E = \dfrac{1}{{\left| x \right| + 2017}}\\
\left| x \right| \ge 0,\,\,\,\forall x \Rightarrow \left| x \right| + 2017 \ge 2017,\,\,\,\forall x\\
\Rightarrow \dfrac{1}{{\left| x \right| + 2017}} \le \dfrac{1}{{2017}},\,\,\,\forall x\\
\Rightarrow E \le \dfrac{1}{{2017}},\,\,\,\forall x\\
\Rightarrow {E_{\max }} = \dfrac{1}{{2017}} \Leftrightarrow \left| x \right| = 0 \Leftrightarrow x = 0
\end{array}\)
