2) a) Đặt \(\left(x-1\right)\left(x+6\right)=t\)
\(\Leftrightarrow x^2+5x-6=t\)
\(\left(x+2\right)\left(x+3\right)=x^2+5x+6=t+12\)
\(A=t\left(t+12\right)+2042\)
\(A=t^2+12t+2042\)
\(A=\left(t+6\right)^2-6^2+2042\)
\(A=\left(t+6\right)^2+2006\)
\(\left(t+6\right)^2\ge0\Rightarrow\left(t+6\right)^2+2006\ge2006\)
\(Min_A=2006\) khi \(\left(t+6\right)^2=0\Leftrightarrow t=-6\Leftrightarrow x^2+5x-6=-6\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy: MinA=2006 khi x=0 hoặc x=5
Bài 2b làm tương tự