\(36x^2+16y^2=9\)
Ta có:
\(\left(y-2x\right)^2\le\left(\left(\frac{1}{4}\right)^2+\left(\frac{1}{3}\right)^2\right)\left(\left(4y\right)^2+\left(-6x\right)^2\right)\)
\(\Rightarrow\left(y-2x\right)^2\le\frac{25}{144}.9=\frac{25}{16}\)
\(\Rightarrow\frac{-5}{4}\le y-2x\le\frac{5}{4}\Rightarrow\frac{15}{4}\le y-2x+5\le\frac{25}{4}\)