Để mik suy nghĩ đã sau đó mik trả lời giúp bạn nhé!
\(x^2-4xy+4y^2+3x^2-2x+\frac{1}{3}-\frac{1}{3}\\ =\left(x-2y\right)^2+3\left(x-\frac{1}{3}\right)^2-\frac{1}{3}\ge-\frac{1}{3}\)
khi \(x=\frac{1}{3},y=\frac{1}{6}\)
Ta có:
\(4x^2+4y^2−4xy−2x\) = \(x^2-4xy+4y^2+2x^2+x^2-2x+1-1\)
=\(\left(x-2y\right)^2+2x^2+\left(x-1\right)^2-1\)
Vì \((x-2y)^2\ge0\);\(2x^2\ge0\);\((x-1)^2\ge0\)
\(\Rightarrow\left(x-2y\right)^2+2x^2+\left(x-1\right)^2-1\ge-1\)
Min 4x2+4y2−4xy−2x là -1 khi \(\hept{\begin{cases}x-2y=0\\x-1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\end{cases}}\)
