Đáp án:

B2:

c. \(Min = 4\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
a.4{a^2} - 8x + 25 = 4{a^2} - 2.2a.2 + 4 + 21\\
 = {\left( {2a - 2} \right)^2} + 21\\
Do:{\left( {2a - 2} \right)^2} \ge 0\forall a \in R\\
 \to {\left( {2a - 2} \right)^2} + 21 > 0\forall a \in R\\
b. - {a^2} + 10a - 35 =  - \left( {{a^2} - 10a + 35} \right)\\
 =  - \left( {{a^2} - 2.a.5 + 25 + 10} \right)\\
 =  - {\left( {a - 5} \right)^2} - 10\\
Do:{\left( {a - 5} \right)^2} \ge 0\forall a \in R\\
 \to  - {\left( {a - 5} \right)^2} \le 0\forall a \in R\\
 \to  - {\left( {a - 5} \right)^2} - 10 < 0\forall a \in R\\
B2:\\
a.M = {x^2} - 12x + 33\\
 = {x^2} - 2.x.6 + 36 - 3\\
 = {\left( {x - 6} \right)^2} - 3\\
Do:{\left( {x - 6} \right)^2} \ge 0\forall x \in R\\
 \to {\left( {x - 6} \right)^2} - 3 \ge  - 3\\
 \to Min =  - 3\\
 \Leftrightarrow x - 6 = 0\\
 \Leftrightarrow x = 6\\
b.N = {x^2} + 2x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{{11}}{4}\\
 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4}\\
Do:{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0\forall x \in R\\
 \to {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4} \ge \dfrac{{11}}{4}\\
 \to Min = \dfrac{{11}}{4}\\
 \Leftrightarrow x + \dfrac{1}{2} = 0\\
 \Leftrightarrow x =  - \dfrac{1}{2}\\
c.P = 9{x^2} - 2.3x.1 + 1 + 4\\
 = {\left( {3x - 1} \right)^2} + 4\\
Do:{\left( {3x - 1} \right)^2} \ge 0\forall x \in R\\
 \to {\left( {3x - 1} \right)^2} + 4 \ge 4\\
 \to Min = 4\\
 \Leftrightarrow x = \dfrac{1}{3}
\end{array}\)