Toán Lớp 9: Tìm GTLN của biểu thức sau: Q=-x+4 căn x+1
M =-4x + 6 căn x -2
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Toán Lớp 9: Tìm GTLN của biểu thức sau: Q=-x+4 căn x+1
M =-4x + 6 căn x -2
Comments ( 1 )
MaxQ = 5\\
MaxM = \dfrac{1}{4}
\end{array}\)
DK:x \ge 0\\
Q = – x + 4\sqrt x + 1\\
= – \left( {x – 4\sqrt x – 1} \right)\\
= – \left( {x – 4\sqrt x + 4 – 5} \right)\\
= – {\left( {\sqrt x – 2} \right)^2} + 5\\
Do:{\left( {\sqrt x – 2} \right)^2} \ge 0\forall x \ge 0\\
\to – {\left( {\sqrt x – 2} \right)^2} \le 0\\
\to – {\left( {\sqrt x – 2} \right)^2} + 5 \le 5\\
\to MaxQ = 5\\
\Leftrightarrow \sqrt x – 2 = 0\\
\to x = 4\\
M = – 4x + 6\sqrt x – 2\\
= – \left( {4x – 6\sqrt x + 2} \right)\\
= – \left( {4x – 2.2\sqrt x .\dfrac{3}{2} + \dfrac{9}{4} – \dfrac{1}{4}} \right)\\
= – {\left( {2\sqrt x – \dfrac{3}{2}} \right)^2} + \dfrac{1}{4}\\
Do:{\left( {2\sqrt x – \dfrac{3}{2}} \right)^2} \ge 0\forall x \ge 0\\
\to – {\left( {2\sqrt x – \dfrac{3}{2}} \right)^2} \le 0\\
\to – {\left( {2\sqrt x – \dfrac{3}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
\to MaxM = \dfrac{1}{4}\\
\Leftrightarrow 2\sqrt x – \dfrac{3}{2} = 0\\
\to x = \dfrac{9}{{16}}
\end{array}\)