Đáp án:
\[{A_{\min }} = 2020 \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 2\\
z = 1
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = {x^2} + {y^2} + {z^2} - yz - 4x - 3y + 2027\\
= \left( {{x^2} - 4x + 4} \right) + \left( {\dfrac{3}{4}{y^2} - 3y + 3} \right) + \left( {\dfrac{1}{4}{y^2} - yz + {z^2}} \right) + 2020\\
= \left( {{x^2} - 2.x.2 + {2^2}} \right) + 3.\left( {\dfrac{1}{4}{y^2} - y + 1} \right) + \left[ {{{\left( {\dfrac{1}{2}y} \right)}^2} - 2.\dfrac{1}{2}y.z + {z^2}} \right] + 2020\\
= {\left( {x - 2} \right)^2} + 3.\left[ {{{\left( {\dfrac{1}{2}y} \right)}^2} - 2.\dfrac{1}{2}y.1 + 1} \right] + {\left( {\dfrac{1}{2}y - z} \right)^2} + 2020\\
= {\left( {x - 2} \right)^2} + 3.{\left( {\dfrac{1}{2}y - 1} \right)^2} + {\left( {\dfrac{1}{2}y - z} \right)^2} + 2020\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {\dfrac{1}{2}y - 1} \right)^2} \ge 0,\,\,\,\forall y\\
{\left( {\dfrac{1}{2}y - z} \right)^2} \ge 0,\,\,\,\forall y,z\\
\Rightarrow A = {\left( {x - 2} \right)^2} + 3.{\left( {\dfrac{1}{2}y - 1} \right)^2} + {\left( {\dfrac{1}{2}y - z} \right)^2} + 2020 \ge 2020,\,\,\,\forall x,y,z\\
\Rightarrow {A_{\min }} = 2020 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 2} \right)^2} = 0\\
{\left( {\dfrac{1}{2}y - 1} \right)^2} = 0\\
{\left( {\dfrac{1}{2}y - z} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - 2 = 0\\
\dfrac{1}{2}y - 1 = 0\\
\dfrac{1}{2}y - z = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 2\\
z = 1
\end{array} \right.
\end{array}\)
Vậy \({A_{\min }} = 2020 \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 2\\
z = 1
\end{array} \right.\)
