1) \(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=8\left(x+4\right)\left(x-1\right)\)
2) \(A=x^2+2y^2+2xy-2y+2021=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2020=\left(x+y\right)^2+\left(y-1\right)^2+2020\ge2020\)
\(minA=2020\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=2.\left(x+4\right).4\left(x-1\right)=8\left(x-1\right)\left(x+4\right)\)
Bài 1:
Ta có: \(\left(3x+2\right)^2-\left(x-6\right)^2\)
\(=\left(3x+2-x+6\right)\left(3x+2+x-6\right)\)
\(=\left(2x+8\right)\left(4x-4\right)\)
\(=8\left(x+4\right)\left(x-1\right)\)
