Tìm GTNN của: A= 3x ²+18+33 B= x ² -6x+10+y ² C= (2x-1) ² +(x+2) ² D= $\frac{-2}{7x²-8x+7}$ 09/08/2021 Bởi Gianna Tìm GTNN của: A= 3x ²+18+33 B= x ² -6x+10+y ² C= (2x-1) ² +(x+2) ² D= $\frac{-2}{7x²-8x+7}$
Đáp án: b. \(Min = 1\) Giải thích các bước giải: \(\begin{array}{l}A = 3{x^2} + 18x + 33\\ = {\left( {x\sqrt 3 } \right)^2} + 2.x\sqrt 3 .\dfrac{9}{{\sqrt 3 }} + {\left( {\dfrac{9}{{\sqrt 3 }}} \right)^2} + 6\\ = {\left( {x\sqrt 3 + \dfrac{9}{{\sqrt 3 }}} \right)^2} + 6\\Do:{\left( {x\sqrt 3 + \dfrac{9}{{\sqrt 3 }}} \right)^2} \ge 0\forall x \in R\\ \to {\left( {x\sqrt 3 + \dfrac{9}{{\sqrt 3 }}} \right)^2} + 6 \ge 6\\ \to Min = 6\\ \Leftrightarrow x\sqrt 3 + \dfrac{9}{{\sqrt 3 }} = 0\\ \Leftrightarrow x = – 3\\B = {x^2} – 6x + 10 + {y^2}\\ = \left( {{x^2} – 2.3.x + 9 + 1} \right) + {y^2}\\ = {\left( {x – 3} \right)^2} + {y^2} + 1\\Do:\left\{ \begin{array}{l}{\left( {x – 3} \right)^2} \ge 0\forall x \in R\\{y^2} \ge 0\forall y \in R\end{array} \right.\\ \to {\left( {x – 3} \right)^2} + {y^2} \ge 0\\ \to {\left( {x – 3} \right)^2} + {y^2} + 1 \ge 1\\ \to Min = 1\\ \Leftrightarrow \left\{ \begin{array}{l}x = 3\\y = 0\end{array} \right.\\C = {\left( {2x – 1} \right)^2} + {\left( {x + 2} \right)^2}\\Do:\left\{ \begin{array}{l}{\left( {2x – 1} \right)^2} \ge 0\forall x\\{\left( {x + 2} \right)^2} \ge 0\forall x\end{array} \right.\\ \to {\left( {2x – 1} \right)^2} + {\left( {x + 2} \right)^2} \ge 0\\ \to Min = 0\\ \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{1}{2}\\x = – 2\end{array} \right.\left( {vô lý} \right)\end{array}\) ⇒ Không tồn tại giá trị x để C đạt GTNN \(\begin{array}{l}D = \dfrac{{ – 2}}{{7{x^2} – 8x + 7}}\\ = \dfrac{{ – 2}}{{{{\left( {x\sqrt 7 } \right)}^2} – 2.x\sqrt 7 .\dfrac{4}{{\sqrt 7 }} + {{\left( {\dfrac{4}{{\sqrt 7 }}} \right)}^2} + \dfrac{{33}}{7}}}\\ = \dfrac{{ – 2}}{{{{\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)}^2} + \dfrac{{33}}{7}}}\\Do:{\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)^2} \ge 0\forall x\\ \to {\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)^2} + \dfrac{{33}}{7} \ge \dfrac{{33}}{7}\\ \to \dfrac{2}{{{{\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)}^2} + \dfrac{{33}}{7}}} \le \dfrac{{14}}{{33}}\\ \to – \dfrac{2}{{{{\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)}^2} + \dfrac{{33}}{7}}} \ge – \dfrac{{14}}{{33}}\\ \to Min = – \dfrac{{14}}{{33}}\\ \Leftrightarrow x\sqrt 7 – \dfrac{4}{{\sqrt 7 }} = 0\\ \Leftrightarrow x = \dfrac{4}{7}\end{array}\) Bình luận
Đáp án:
b. \(Min = 1\)
Giải thích các bước giải:
\(\begin{array}{l}
A = 3{x^2} + 18x + 33\\
= {\left( {x\sqrt 3 } \right)^2} + 2.x\sqrt 3 .\dfrac{9}{{\sqrt 3 }} + {\left( {\dfrac{9}{{\sqrt 3 }}} \right)^2} + 6\\
= {\left( {x\sqrt 3 + \dfrac{9}{{\sqrt 3 }}} \right)^2} + 6\\
Do:{\left( {x\sqrt 3 + \dfrac{9}{{\sqrt 3 }}} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x\sqrt 3 + \dfrac{9}{{\sqrt 3 }}} \right)^2} + 6 \ge 6\\
\to Min = 6\\
\Leftrightarrow x\sqrt 3 + \dfrac{9}{{\sqrt 3 }} = 0\\
\Leftrightarrow x = – 3\\
B = {x^2} – 6x + 10 + {y^2}\\
= \left( {{x^2} – 2.3.x + 9 + 1} \right) + {y^2}\\
= {\left( {x – 3} \right)^2} + {y^2} + 1\\
Do:\left\{ \begin{array}{l}
{\left( {x – 3} \right)^2} \ge 0\forall x \in R\\
{y^2} \ge 0\forall y \in R
\end{array} \right.\\
\to {\left( {x – 3} \right)^2} + {y^2} \ge 0\\
\to {\left( {x – 3} \right)^2} + {y^2} + 1 \ge 1\\
\to Min = 1\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 0
\end{array} \right.\\
C = {\left( {2x – 1} \right)^2} + {\left( {x + 2} \right)^2}\\
Do:\left\{ \begin{array}{l}
{\left( {2x – 1} \right)^2} \ge 0\forall x\\
{\left( {x + 2} \right)^2} \ge 0\forall x
\end{array} \right.\\
\to {\left( {2x – 1} \right)^2} + {\left( {x + 2} \right)^2} \ge 0\\
\to Min = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
x = – 2
\end{array} \right.\left( {vô lý} \right)
\end{array}\)
⇒ Không tồn tại giá trị x để C đạt GTNN
\(\begin{array}{l}
D = \dfrac{{ – 2}}{{7{x^2} – 8x + 7}}\\
= \dfrac{{ – 2}}{{{{\left( {x\sqrt 7 } \right)}^2} – 2.x\sqrt 7 .\dfrac{4}{{\sqrt 7 }} + {{\left( {\dfrac{4}{{\sqrt 7 }}} \right)}^2} + \dfrac{{33}}{7}}}\\
= \dfrac{{ – 2}}{{{{\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)}^2} + \dfrac{{33}}{7}}}\\
Do:{\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)^2} + \dfrac{{33}}{7} \ge \dfrac{{33}}{7}\\
\to \dfrac{2}{{{{\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)}^2} + \dfrac{{33}}{7}}} \le \dfrac{{14}}{{33}}\\
\to – \dfrac{2}{{{{\left( {x\sqrt 7 – \dfrac{4}{{\sqrt 7 }}} \right)}^2} + \dfrac{{33}}{7}}} \ge – \dfrac{{14}}{{33}}\\
\to Min = – \dfrac{{14}}{{33}}\\
\Leftrightarrow x\sqrt 7 – \dfrac{4}{{\sqrt 7 }} = 0\\
\Leftrightarrow x = \dfrac{4}{7}
\end{array}\)