Câu 6:
\(A=2x^2+4xy+3y^2-2x+2y+2\)
\(\dfrac{A}{2}=x^2+2xy+\dfrac{3}{2}y^2-x+y+1\)
\(=\left(x^2+y^2+\dfrac{1}{4}+2xy-x-y\right)+\left(\dfrac{1}{2}y^2+2y+\dfrac{3}{4}\right)\)
\(=\left(x^2+y^2+2xy\right)-\left(x+y\right)+\dfrac{1}{4}+\dfrac{1}{2}\left(y^2+4y+\dfrac{3}{2}\right)\)
\(=\left(x+y-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\left[\left(y+2\right)^2-\dfrac{5}{2}\right]\)
\(\ge\dfrac{1}{2}.\left(-\dfrac{5}{2}\right)=\dfrac{-5}{4}\)
\(\Rightarrow A\ge\dfrac{-5}{2}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y-\dfrac{1}{2}=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-2\end{matrix}\right.\)
Vậy \(MinA=\dfrac{-5}{2}\)
giúp mình câu b bài 5 vs câu b bài 6 đi
