Đúng là \(\dfrac{x^2}{9}+\dfrac{y^2}{9}=1\) chứ em? Đề thật kì quặc, tại sao ko cho luôn là \(x^2+y^2=9\) cho rồi
Ta có:
\(\left(x+2.y\right)^2\le\left(1+4\right)\left(x^2+y^2\right)=45\)
\(\Rightarrow-3\sqrt{5}\le x+2y\le3\sqrt{5}\)
\(\Rightarrow1-3\sqrt{5}\le x+2y\le1+3\sqrt{5}\)
\(P_{max}=1+3\sqrt{5}\) khi \(\left(x;y\right)=\left(\dfrac{3}{\sqrt{5}};\dfrac{6}{\sqrt{5}}\right)\)
\(P_{min}=1-3\sqrt{5}\) khi \(\left(x;y\right)=\left(-\dfrac{3}{\sqrt{5}};-\dfrac{6}{\sqrt{5}}\right)\)
Nếu đề là:
\(\dfrac{x^2}{9}+\dfrac{y^2}{4}=1\) \(\Leftrightarrow4x^2+9y^2=36\)
Ta có:
\(\left(x+2y\right)^2=\left(\dfrac{1}{2}.2x+\dfrac{2}{3}.3y\right)^2\le\left(\dfrac{1}{4}+\dfrac{4}{9}\right)\left(4x^2+9y^2\right)=25\)
\(\Rightarrow-5\le x+2y\le5\)
\(\Rightarrow-4\le x+2y+1\le6\)
\(P_{max}=6\) khi \(\left(x;y\right)=\left(\dfrac{9}{5};\dfrac{8}{5}\right)\)
\(P_{min}=-4\) khi \(\left(x;y\right)=\left(-\dfrac{9}{5};-\dfrac{8}{5}\right)\)
