Giúp bài 10 nhhhjhhhhhhhhhh

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\,\,\,A = 2{x^2} - 2\\
{x^2} \ge 0 \Rightarrow 2{x^2} \ge 0 \Rightarrow 2{x^2} - 2 \ge  - 2,\,\,\,\forall x\\
 \Rightarrow {A_{\min }} =  - 2 \Leftrightarrow {x^2} = 0 \Leftrightarrow x = 0\\
b,\\
B = \left| {x + \dfrac{1}{3}} \right| - \dfrac{1}{6}\\
\left| {x + \dfrac{1}{3}} \right| \ge 0 \Rightarrow \left| {x + \dfrac{1}{3}} \right| - \dfrac{1}{6} \ge  - \dfrac{1}{6},\,\,\,\forall x\\
 \Rightarrow {B_{\min }} =  - \dfrac{1}{6} \Leftrightarrow \left| {x + \dfrac{1}{3}} \right| = 0 \Leftrightarrow x =  - \dfrac{1}{3}\\
c,\\
C = \dfrac{{\left| x \right| + 2017}}{{2018}}\\
\left| x \right| \ge 0 \Rightarrow \left| x \right| + 2017 \ge 2017 \Rightarrow \dfrac{{\left| x \right| + 2017}}{{2018}} \ge \dfrac{{2017}}{{2018}},\,\,\forall x\\
 \Rightarrow {C_{\min }} = \dfrac{{2017}}{{2018}} \Leftrightarrow \left| x \right| = 0 \Leftrightarrow x = 0\\
d,\\
D = 3 - {\left( {x + 1} \right)^2}\\
{\left( {x + 1} \right)^2} \ge 0 \Leftrightarrow 3 - {\left( {x + 1} \right)^2} \le 3,\,\,\,\forall x\\
 \Rightarrow {D_{\max }} = 3 \Leftrightarrow {\left( {x + 1} \right)^2} = 0 \Leftrightarrow x =  - 1\\
e,\\
E =  - \left| {0,1 + x} \right| - 1,9 =  - \left( {\left| {0,1 + x} \right| + 1,9} \right)\\
\left| {0,1 + x} \right| \ge 0 \Rightarrow \left| {0,1 + x} \right| + 1,9 \ge 1,9,\,\,\,\forall x\\
 \Rightarrow  - \left( {\left| {0,1 + x} \right| + 1,9} \right) \le  - 1,9,\,\,\,\forall x\\
 \Rightarrow {E_{\max }} =  - 1,9 \Leftrightarrow \left| {0,1 + x} \right| = 0 \Leftrightarrow x =  - 0,1\\
f,\\
F = \dfrac{1}{{\left| x \right| + 2017}}\\
\left| x \right| \ge 0 \Leftrightarrow \left| x \right| + 2017 \ge 2017,\,\,\,\forall x\\
 \Rightarrow \dfrac{1}{{\left| x \right| + 2017}} \le \dfrac{1}{{2017}},\,\,\,\forall x\\
 \Rightarrow {F_{\max }} = \dfrac{1}{{2017}} \Leftrightarrow \left| x \right| = 0 \Leftrightarrow x = 0
\end{array}\)