\(A=\dfrac{1}{5x-3\sqrt{x}+5}=\dfrac{1}{5\left(x-\dfrac{3}{5}\sqrt{x}+1\right)}\\ =\dfrac{1}{5\left(x-2.\dfrac{3}{10}\sqrt{x}+\dfrac{9}{100}\right)+\dfrac{91}{20}}=\dfrac{1}{5\left(\sqrt{x}-\dfrac{3}{10}\right)^2+\dfrac{91}{20}}\le\dfrac{20}{91}\)
Vậy GTLN của A là \(\dfrac{20}{91}\) tại \(x=\dfrac{9}{100}\)
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ĐK \(x\ge0\)
Ta có: \(A=\dfrac{1}{5x-3\sqrt{x}+5}=\dfrac{1}{5x-2\cdot5\cdot\dfrac{3}{10}\sqrt{x}+5\cdot\dfrac{9}{100}+\dfrac{91}{20}}\)
\(\Leftrightarrow A=\dfrac{1}{5\left(x-2\cdot\dfrac{3}{10}\sqrt{x}+\dfrac{9}{100}\right)+\dfrac{91}{20}}=\dfrac{1}{5\left(\sqrt{x}-\dfrac{3}{10}\right)^2+\dfrac{91}{20}}\)
\(\Leftrightarrow A\ge\dfrac{1}{\dfrac{91}{20}}=\dfrac{20}{91}\) (vì \(\left(\sqrt{x}-\dfrac{3}{10}\right)^2\ge0\))
Vậy GTNN của A là 20/91 <=> x=9/100
điều kiên xác định \(x\ge0\)
để A đạt giá trị lớn nhất thì \(5x-3\sqrt{x}+5\) bé nhất có
\(5x-3\sqrt{x}+5\\ =\left(\sqrt{5x}\right)^2-2\times\dfrac{3}{50}\sqrt{25x}+\dfrac{9}{2500}+4,9964\\ =\left(\sqrt{5x}-\dfrac{3}{50}\right)^2+4,9964\)
có \(\left(\sqrt{5x}-\dfrac{3}{50}\right)^2\ge0\\ \Rightarrow\left(\sqrt{5x}-\dfrac{3}{50}\right)^2+4,9964\ge4,9964\)
nên A lớn nhất là \(\dfrac{1}{4,9964}=\dfrac{2500}{12491}\)
khi \(\sqrt{5x}-\dfrac{3}{50}=0\\ \Leftrightarrow5x=\dfrac{9}{2500}\\ \Leftrightarrow x=\dfrac{9}{12500}\)
