a,Ta có: \(2A=4x^2+4xy+2y^2-4x+4y+4\)

\(=4x^2+2x\left(y-2\right)+\left(y-2\right)^2+y^2+8y+16-20\)

\(=\left(2x+y-2\right)^2+\left(y+4\right)^2-20\)

Vì \(\left\{{}\begin{matrix}\left(2x+y-2\right)^2\ge0\\\left(y+4\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow2A\ge-20\Rightarrow A\ge-10\)

Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-4\end{matrix}\right.\)

Vậy ....

c,Ta có:\(4C=4x^2+4xy+4y^2-12x-12y\)

\(=4x^2+2.2x\left(y-3\right)+\left(y-3\right)^2-\left(y-3\right)^2+4y^2-12y\)

\(=\left(2x+y-3\right)^2+3\left(y^2-2y+1\right)-12\)

\(=\left(2x+y-3\right)^2+3\left(y-1\right)^2-12\)

Vì \(\left\{{}\begin{matrix}\left(2x+y-3\right)^2\ge0\\3\left(y-1\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow4C\ge-12\Rightarrow C\ge-3\)

Dấu ''='' xảy ra \(\Leftrightarrow x=y=1\)

Vậy ...

a,Ta có:\(B=x^4-x^3y+y^4-xy^3+x^2y^2-8xy+16+184\)

\(=x\left(x^3-y^3\right)-y\left(x^3-y^3\right)+\left(xy-4\right)^2+184\)

\(=\left(x-y\right)^2\left(x^2+xy+y^2\right)+\left(xy-4\right)^2+184\)

\(=\left(x-y\right)^2\left[\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3y^2}{4}\right]+\left(xy-4\right)^2+184\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left[\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3y^2}{4}\right]\ge0\\\left(xy-4\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow B\ge184\)

Dấu ''='' xảy ra \(\Leftrightarrow x=y=2\)

Vậy ...

\(B=\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+6\)

\(B=\left(x-2\right)\left(x-5\right)\left(x-3\right)\left(x-4\right)+6\)

\(B=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+6\)

Đặt \(x^2-7x+11=t\)

\(B=\left(t-1\right)\left(t+1\right)+6=t^2+5\ge5\)