* \(3x+y=1\Rightarrow y=1-3x\)
\(M=3x^2+\left(1-3x\right)^2=3x^2+1-6x+9x^2=12x^2-6x+1=12\left(x^2-\dfrac{1}{2}x+\dfrac{1}{12}\right)=12\left(x^2-2.x.\dfrac{1}{4}+\dfrac{1}{16}\right)+\dfrac{1}{4}=12\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)\(\Rightarrow Min_M=\dfrac{1}{4}\Leftrightarrow x=y=\dfrac{1}{4}\)
\(N=x^2+xy+y^2-3x-3y\)
\(4N=4x^2+4xy+4y^2-12x-12y\)
\(4N=\left(4x^2+4xy+y^2\right)-12x-6y+9+3y^2-6y+3-12\)
\(4N=\left(2x+y\right)^2-2.3\left(2x+y\right)+9+3\left(y-1\right)^2-12\)
\(4N=\left(2x+y-3\right)^2+3\left(y-1\right)^2-10\ge-12\)
\(\Rightarrow N\ge-3\)
\(\Rightarrow Min_N=-3\Leftrightarrow x=y=1\)
Đúng 0
Bình luận (4)