tìm giá trị lớn nhất b)B= căn x phần x+ căn x +1 ( x>0) c)C= căn x phần x – căn x +1 (x>0) 14/08/2021 Bởi Genesis tìm giá trị lớn nhất b)B= căn x phần x+ căn x +1 ( x>0) c)C= căn x phần x – căn x +1 (x>0)
Đáp án: $\begin{array}{l}b)B = \dfrac{{\sqrt x }}{{x + \sqrt x + 1}}\left( {x > 0} \right)\\ \Rightarrow B.x + B.\sqrt x + B = \sqrt x \\ \Rightarrow B.x + \left( {B – 1} \right).\sqrt x + B = 0\\ \Rightarrow \left\{ \begin{array}{l}\dfrac{{1 – B}}{B} \ge 0\\\Delta \ge 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}0 \le B \le 1\\{B^2} – 2B + 1 – 4{B^2} \ge 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}0 \le B \le 1\\3{B^2} + 2B – 1 \le 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}0 \le B \le 1\\\left( {3B – 1} \right)\left( {B + 1} \right) \le 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}0 \le B \le 1\\ – 1 \le B \le \dfrac{1}{3}\end{array} \right.\\ \Rightarrow 0 \le B \le \dfrac{1}{3}\\ \Rightarrow GTLN:B = \dfrac{1}{3}\,khi:x = 1\\c)C = \dfrac{{\sqrt x }}{{x – \sqrt x + 1}}\\ \Rightarrow C.x – C.\sqrt x + C = \sqrt x \\ \Rightarrow C.x – \left( {C + 1} \right).\sqrt x + C = 0\\ \Rightarrow \left\{ \begin{array}{l}C > 0\\\Delta \ge 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}C > 0\\{\left( {C + 1} \right)^2} – 4{C^2} \ge 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}C > 0\\{C^2} + 2C + 1 – 4{C^2} \ge 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}C > 0\\3{C^2} – 2C – 1 \le 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}C > 0\\\left( {3C + 1} \right)\left( {C – 1} \right) \le 0\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}C > 0\\ – \dfrac{1}{3} \le C \le 1\end{array} \right.\\ \Rightarrow 0 < C \le 1\\ \Rightarrow GTLN:C = 1\,khi:x = 1\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
b)B = \dfrac{{\sqrt x }}{{x + \sqrt x + 1}}\left( {x > 0} \right)\\
\Rightarrow B.x + B.\sqrt x + B = \sqrt x \\
\Rightarrow B.x + \left( {B – 1} \right).\sqrt x + B = 0\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{1 – B}}{B} \ge 0\\
\Delta \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 \le B \le 1\\
{B^2} – 2B + 1 – 4{B^2} \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 \le B \le 1\\
3{B^2} + 2B – 1 \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 \le B \le 1\\
\left( {3B – 1} \right)\left( {B + 1} \right) \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 \le B \le 1\\
– 1 \le B \le \dfrac{1}{3}
\end{array} \right.\\
\Rightarrow 0 \le B \le \dfrac{1}{3}\\
\Rightarrow GTLN:B = \dfrac{1}{3}\,khi:x = 1\\
c)C = \dfrac{{\sqrt x }}{{x – \sqrt x + 1}}\\
\Rightarrow C.x – C.\sqrt x + C = \sqrt x \\
\Rightarrow C.x – \left( {C + 1} \right).\sqrt x + C = 0\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
\Delta \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
{\left( {C + 1} \right)^2} – 4{C^2} \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
{C^2} + 2C + 1 – 4{C^2} \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
3{C^2} – 2C – 1 \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
\left( {3C + 1} \right)\left( {C – 1} \right) \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
C > 0\\
– \dfrac{1}{3} \le C \le 1
\end{array} \right.\\
\Rightarrow 0 < C \le 1\\
\Rightarrow GTLN:C = 1\,khi:x = 1
\end{array}$