\(K=5x^2+4xy+y\left(y-4\right)-10x\)
\(K=5x^2+4xy+y^2-4y-10x\)
\(K=\left(4x^2+4xy+y^2\right)+x^2-4y-10x\)
\(K=\left[\left(2x+y\right)^2-2\left(2x+y\right).2+4\right]+\left(x^2-2x+1\right)-5\)
\(K=\left(2x+y-2\right)^2+\left(x-1\right)^2-5\)
Mà \(\left(2x+y-2\right)^2\ge0\forall x;y\)
\(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow K\ge-5\)
Dấu "=" xảy ra khi : \(\hept{\begin{cases}2x+y-2=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=0\\x=1\end{cases}}\)
Vậy \(K_{Min}=-5\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)
\(K=5x^2+4xy+y\left(y-4\right)-10x.\)
\(=\left(4x^2+y^2+4+4xy-8x-4y\right)+\left(x^2-2x+1\right)-1\)
\(=\left(\left(2x\right)^2+y^2+2^2+2.2x.y-2.2x.2-2.y.2\right)+\left(x^2-2x+1\right)-1\)
\(=\left(2x+y-2\right)^2+\left(x-1\right)^2-1\ge-1\)
Dấu "=" xảy ra khi
\(\hept{\begin{cases}\left(2x+y-2\right)^2=0\\\left(x-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=0\end{cases}}}\)
Câu trên T làm sai rồi. Quên để ý phía trước có cộng thêm 4. MinK=-5
