Đáp án:

\(\begin{array}{l}
9,\\
{K_{\min }} = \dfrac{3}{4} \Leftrightarrow x = \dfrac{1}{2}\\
10,\\
{M_{\min }} = \dfrac{3}{4} \Leftrightarrow x =  - \dfrac{1}{2}\\
11,\\
{N_{\min }} =  - \dfrac{1}{4} \Leftrightarrow x =  - \dfrac{1}{2}\\
12,\\
{L_{\min }} =  - \dfrac{1}{4} \Leftrightarrow x = \dfrac{1}{2}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
9,\\
K = {x^2} - x + 1\\
 = \left( {{x^2} - x + \dfrac{1}{4}} \right) + \dfrac{3}{4}\\
 = \left[ {{x^2} - 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] + \dfrac{3}{4}\\
 = {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow K = {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4},\,\,\,\forall x\\
 \Rightarrow {K_{\min }} = \dfrac{3}{4} \Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x - \dfrac{1}{2} = 0 \Leftrightarrow x = \dfrac{1}{2}\\
 \Rightarrow {K_{\min }} = \dfrac{3}{4} \Leftrightarrow x = \dfrac{1}{2}\\
10,\\
M = {x^2} + x + 1\\
 = \left( {{x^2} + x + \dfrac{1}{4}} \right) + \dfrac{3}{4}\\
 = \left[ {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] + \dfrac{3}{4}\\
 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow M = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4},\,\,\,\forall x\\
 \Rightarrow {M_{\min }} = \dfrac{3}{4} \Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x + \dfrac{1}{2} = 0 \Leftrightarrow x =  - \dfrac{1}{2}\\
 \Rightarrow {M_{\min }} = \dfrac{3}{4} \Leftrightarrow x =  - \dfrac{1}{2}\\
11,\\
N = {x^2} + x\\
 = \left( {{x^2} + x + \dfrac{1}{4}} \right) - \dfrac{1}{4}\\
 = \left[ {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] - \dfrac{1}{4}\\
 = {\left( {x + \dfrac{1}{2}} \right)^2} - \dfrac{1}{4}\\
{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow N = {\left( {x + \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge  - \dfrac{1}{4},\,\,\,\forall x\\
 \Rightarrow {N_{\min }} =  - \dfrac{1}{4} \Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x + \dfrac{1}{2} = 0 \Leftrightarrow x =  - \dfrac{1}{2}\\
 \Rightarrow {N_{\min }} =  - \dfrac{1}{4} \Leftrightarrow x =  - \dfrac{1}{2}\\
12,\\
L = {x^2} - x\\
 = \left( {{x^2} - x + \dfrac{1}{4}} \right) - \dfrac{1}{4}\\
 = \left[ {{x^2} - 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] - \dfrac{1}{4}\\
 = {\left( {x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4}\\
{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow L = {\left( {x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge  - \dfrac{1}{4},\,\,\,\forall x\\
 \Rightarrow {L_{\min }} =  - \dfrac{1}{4} \Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x - \dfrac{1}{2} = 0 \Leftrightarrow x = \dfrac{1}{2}\\
 \Rightarrow {L_{\min }} =  - \dfrac{1}{4} \Leftrightarrow x = \dfrac{1}{2}
\end{array}\)