Đáp án:

\(\begin{array}{l}
1,\\
a,\\
{A_{\min }} = 5 \Leftrightarrow x =  - 5\\
b,\\
{B_{\min }} = \dfrac{{79}}{4} \Leftrightarrow x =  - \dfrac{5}{4}\\
c,\\
{C_{\min }} = \dfrac{{111}}{4} \Leftrightarrow x =  - \dfrac{1}{2}\\
2,\\
a,\\
{A_{\max }} = 55 \Leftrightarrow x = 5\\
b,\\
{B_{\max }} = \dfrac{{33}}{4} \Leftrightarrow x = \dfrac{5}{4}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
A = {x^2} + 10x + 30 = \left( {{x^2} + 10x + 25} \right) + 5\\
 = \left( {{x^2} + 2.x.5 + {5^2}} \right) + 5 = {\left( {x + 5} \right)^2} + 5\\
{\left( {x + 5} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow A = {\left( {x + 5} \right)^2} + 5 \ge 5,\,\,\,\,\forall x\\
 \Rightarrow {A_{\min }} = 5 \Leftrightarrow {\left( {x + 5} \right)^2} = 0 \Leftrightarrow x + 5 = 0 \Leftrightarrow x =  - 5\\
b,\\
B = 4{x^2} + 10x + 26 = \left( {4{x^2} + 10x + \dfrac{{25}}{4}} \right) + \dfrac{{79}}{4}\\
 = \left[ {{{\left( {2x} \right)}^2} + 2.2x.\dfrac{5}{2} + {{\left( {\dfrac{5}{2}} \right)}^2}} \right] + \dfrac{{79}}{4} = {\left( {2x + \dfrac{5}{2}} \right)^2} + \dfrac{{79}}{4}\\
{\left( {2x + \dfrac{5}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow B = {\left( {2x + \dfrac{5}{2}} \right)^2} + \dfrac{{79}}{4} \ge \dfrac{{79}}{4},\,\,\,\,\forall x\\
 \Rightarrow {B_{\min }} = \dfrac{{79}}{4} \Leftrightarrow {\left( {2x + \dfrac{5}{2}} \right)^2} = 0 \Leftrightarrow 2x + \dfrac{5}{2} = 0 \Leftrightarrow x =  - \dfrac{5}{4}\\
c,\\
C = 9{x^2} + 9x + 30 = \left( {9{x^2} + 9x + \dfrac{9}{4}} \right) + \dfrac{{111}}{4}\\
 = \left[ {{{\left( {3x} \right)}^2} + 2.3x.\dfrac{3}{2} + {{\left( {\dfrac{3}{2}} \right)}^2}} \right] + \dfrac{{111}}{4} = {\left( {3x + \dfrac{3}{2}} \right)^2} + \dfrac{{111}}{4}\\
{\left( {3x + \dfrac{3}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow C = {\left( {3x + \dfrac{3}{2}} \right)^2} + \dfrac{{111}}{4} \ge \dfrac{{111}}{4},\,\,\,\forall x\\
 \Rightarrow {C_{\min }} = \dfrac{{111}}{4} \Leftrightarrow {\left( {3x + \dfrac{3}{2}} \right)^2} = 0 \Leftrightarrow 3x + \dfrac{3}{2} = 0 \Leftrightarrow x =  - \dfrac{1}{2}\\
2,\\
a,\\
A =  - {x^2} + 10x + 30 = 55 + \left( { - {x^2} + 10x - 25} \right)\\
 = 55 - \left( {{x^2} - 10x + 25} \right) = 55 - \left( {{x^2} - 2.x.5 + {5^2}} \right)\\
 = 55 - {\left( {x - 5} \right)^2}\\
{\left( {x - 5} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow A = 55 - {\left( {x - 5} \right)^2} \le 55,\,\,\,\,\forall x\\
 \Rightarrow {A_{\max }} = 55 \Leftrightarrow {\left( {x - 5} \right)^2} = 0 \Leftrightarrow x - 5 = 0 \Leftrightarrow x = 5\\
b,\\
B =  - 4{x^2} - 10x + 2 = \dfrac{{33}}{4} - 4{x^2} - 10x - \dfrac{{25}}{4}\\
 = \dfrac{{33}}{4} - \left( {4{x^2} + 10x + \dfrac{{25}}{4}} \right) = \dfrac{{33}}{4} - \left[ {{{\left( {2x} \right)}^2} - 2.2x.\dfrac{5}{2} + {{\left( {\dfrac{5}{2}} \right)}^2}} \right]\\
 = \dfrac{{33}}{4} - {\left( {2x - \dfrac{5}{2}} \right)^2}\\
{\left( {2x - \dfrac{5}{2}} \right)^2} \ge 0,\,\,\,\,\forall x\\
 \Rightarrow B = \dfrac{{33}}{4} - {\left( {2x - \dfrac{5}{2}} \right)^2} \le \dfrac{{33}}{4},\,\,\,\,\forall x\\
 \Rightarrow {B_{\max }} = \dfrac{{33}}{4} \Leftrightarrow {\left( {2x - \dfrac{5}{2}} \right)^2} = 0 \Leftrightarrow 2x - \dfrac{5}{2} = 0 \Leftrightarrow x = \dfrac{5}{4}
\end{array}\)