-8x2+4xy-y2+10=10-(4x2-4xy+y2)-4x2=10-(2x-y)2-(2x)2
vi-(2x-y)2-(2x)2 ≤0
=>10-(2x-y)2-(2x)2≤10
dau bang say ra khi (2x-y)2-(2x)2=0
vậy gái trị nhỏ nhất là:10
\(Q=-8x^2+4xy-y^2+10\)<=>\(Q=10-4x^2+4xy-y^2-4x^2\)
<=>\(Q=10-\left[\left(2x^2\right)-4xy+y^2\right]-\left(2x\right)^2\)<=>\(Q=10-\left(2x-y\right)^2-\left(2x\right)^2\)
<=>\(Q=10-\left[\left(2x-y\right)^2+\left(2x\right)^2\right]\)
Vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\\left(2x\right)^2\ge0\end{cases}\Leftrightarrow\left(2x-y\right)^2+\left(2x\right)^2\ge0}\)\(\Leftrightarrow-\left[\left(2x-y\right)^2+\left(2x\right)^2\right]\le0\)
\(\Leftrightarrow Q=10-\left[\left(2x-y\right)^2+\left(2x\right)^2\right]\le10\)
=>Qmax=10 <=> \(\left(2x-y\right)^2=\left(2x\right)^2=0\)<=>\(2x-y=2x=0\) <=>\(x=y=0\)
Vậy Qmax=10 tại x=y=0