\(B=3x^2-y+2y^2+x-11=3\left(x+\dfrac{1}{6}\right)^2+2\left(y-\dfrac{1}{4}\right)^2-\dfrac{269}{24}\ge-\dfrac{269}{24}\)
\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=\dfrac{1}{4}\end{matrix}\right.\)
Ta có: \(B=3x^2+x+2y^2-y-11\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}\right)+2\cdot\left(y^2-2\cdot y\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)-\dfrac{269}{24}\)
\(=3\left(x+\dfrac{1}{6}\right)^2+2\left(y-\dfrac{1}{4}\right)^2-\dfrac{269}{24}\ge-\dfrac{269}{24}\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(-\dfrac{1}{6};\dfrac{1}{4}\right)\)
