a,Từ \(3x+y=1\Rightarrow x=\dfrac{1-y}{3}\)
\(\Rightarrow M=3x^2+y^2=3.\left(\dfrac{1-y}{3}\right)^2+y^2=3.\dfrac{y^2-2y+1}{9}+y^2\)
\(=\dfrac{3y^2+y^2-2y+1}{3}=\dfrac{4y^2-2y+1}{3}\)
Ta có: \(4y^2-2y+1=4y^2-2.2y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(2y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Bn tự chứng minh \(\left(2y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2y-\dfrac{1}{2}\right)^2=0\Leftrightarrow y=\dfrac{1}{4}\)
\(\Rightarrow M=\dfrac{\left(2y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{3}\ge\dfrac{\dfrac{3}{4}}{3}=\dfrac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\) \(y=\dfrac{1}{4}\);\(x=\dfrac{1-\dfrac{1}{4}}{3}=\dfrac{1}{4}\)
Cx như a, \(x=\dfrac{1-y}{3}\) thay vào N đc:
\(N=xy=\dfrac{1-y}{3}.y=\dfrac{y-y^2}{3}=\dfrac{-\left(y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+\dfrac{3}{4}\right)}{3}\)
\(=\dfrac{-\left(y-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{3}\)
Bn tự chứng minh \(-\left(y-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\). Dấu "=" xảy ra \(\Leftrightarrow\)\(-\left(y-\dfrac{1}{2}\right)=0\Leftrightarrow y=\dfrac{1}{2}\)
\(\Rightarrow N=\dfrac{-\left(y-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{3}\le\dfrac{\dfrac{-3}{4}}{3}=\dfrac{-1}{4}\)
Vậy MAX N = \(\dfrac{-1}{4}\Leftrightarrow y=\dfrac{1}{2}\)
