Đáp án:
Bài 28:
G. \(Min = 3\)
Giải thích các bước giải:
\(\begin{array}{l}
B27:\\
c.A\left( x \right) = {x^3} - 3\left( {x + 2} \right) + {x^2} - {x^3} + 4x + 9\\
= - 3x - 6 + {x^2} + 4x + 9\\
= {x^2} + x + 3 = {x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{{11}}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4}\\
Do:{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4} > 0\forall x \in R
\end{array}\)
⇒ A(x) = 0 vô nghiệm
\(\begin{array}{l}
B28:\\
G = {\left( {x + y} \right)^2} + {\left( {y - 2} \right)^2} + 3\\
Do:\left\{ \begin{array}{l}
{\left( {x + y} \right)^2} \ge 0\forall x,y \in R\\
{\left( {y - 2} \right)^2} \ge 0\forall x \in R
\end{array} \right.\\
\to {\left( {x + y} \right)^2} + {\left( {y - 2} \right)^2} \ge 0\forall x,y \in R\\
\to {\left( {x + y} \right)^2} + {\left( {y - 2} \right)^2} + 3 \ge 3\\
\to Min = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
x + y = 0\\
y - 2 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 2\\
x = - 2
\end{array} \right.
\end{array}\)
