Đáp án + giải thích các bước giải:
Áp dụng bất đẳng thức Bunhiacopxki
`(1/16+1/9)(16y^2+36x^2)>=[1/4 . 4y+1/3 . (-6x)]^2`
`->25/144 . 9>=(y-2x)^2`
`->(y-2x)^2-(5/4)^2<=0`
`->(y-2x-5/4)(y-2x+5/4)<=0`
`->`\(\left[ \begin{array}{l}\left\{\begin{matrix} y-2x-\dfrac{5}{4}\le0\\y-2x+\dfrac{5}{4}\ge0 \end{matrix}\right.\\\left\{\begin{matrix} y-2x-\dfrac{5}{4}\ge0\\y-2x+\dfrac{5}{4}\le0 \end{matrix}\right.\end{array} \right.\)
`->`\(\left[ \begin{array}{l}\left\{\begin{matrix} y-2x\le \dfrac{5}{4}\\y-2x\ge \dfrac{-5}{4} \end{matrix}\right.\\\left\{\begin{matrix} y-2x\ge \dfrac{5}{4}\\y-2x\le \dfrac{-5}{4} \end{matrix}\right.\end{array} \right.\)
`->-5/4 <= y-2x <= 5/4`
`->15/4<=A<=25/4`
Vậy `A` đạt giá trị lớn nhất là `25/4` khi $ \left\{\begin{matrix} \dfrac{4y}{\dfrac{1}{4}}=\dfrac{-6x}{\dfrac{1}{3}}\\y-2x=\dfrac{5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} 16y=-18x\\y-2x=\dfrac{5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{-9}{8}x\\\dfrac{-9}{8}x-2x=\dfrac{5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{9}{20}\\x=\dfrac{-2}{5}\end{matrix}\right. $
`A` đạt giá trị nhỏ nhất là `15/4` khi $ \left\{\begin{matrix} \dfrac{4y}{\dfrac{1}{4}}=\dfrac{-6x}{\dfrac{1}{3}}\\y-2x=\dfrac{-5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} 16y=-18x\\y-2x=\dfrac{-5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{-9}{8}x\\\dfrac{-9}{8}x-2x=\dfrac{-5}{4} \end{matrix}\right. \\ \rightarrow \left\{\begin{matrix} y=\dfrac{-9}{20}\\x=\dfrac{2}{5}\end{matrix}\right. $
