Ta có: \(A=\left|x^2+x+1\right|+\left|x^2+x-12\right|\)
\(=\left|x^2+x+1\right|+\left|12-x^2-x\right|\)
\(\Leftrightarrow A=\left|x^2+x+1\right|+\left|12-x^2-x\right|\ge\left|x^2+x+1+12-x^2-x\right|=\left|13\right|=13\)
Dấu '=' xảy ra khi
\(\left(x^2+x+1\right)\left(12-x^2-x\right)=0\)
\(\Leftrightarrow12-x^2-x=0\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow x^2+4x-3x-12=0\)
\(\Leftrightarrow x\left(x+4\right)-3\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=3\end{matrix}\right.\)
Vậy: Giá trị nhỏ nhất của biểu thức \(A=\left|x^2+x+1\right|+\left|x^2+x-12\right|\) là 13 khi x∈{-4;3}
