Ta có :
\(3x+y=1\)
\(\Rightarrow y=1-3x\left(1\right)\)
\(A=3x^2+y^2\)
\(\Rightarrow A=3x^2+\left(1-3x\right)^2\left(do\left(1\right)\right)\)
\(\Rightarrow A=3x^2+1-6x+9x^2\)
\(\Rightarrow A=12x^2-6x+1\)
\(\Rightarrow A=12\left(x^2-\dfrac{1}{2}x\right)+1\)
\(\Rightarrow A=12\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{3}{4}+1\)
\(\Rightarrow A=12\left(x-\dfrac{1}{4}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\left(12\left(x-\dfrac{1}{4}\right)^2\ge0,\forall x\in R\right)\)
\(\Rightarrow A\left(min\right)=\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{4}=0\\y=1-3x\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\end{matrix}\right.\) hay \(x=y=\dfrac{1}{4}\)
