Question

tìm gtnn của |x^2+x+1| +|x^2+x-12|

Answer 1

Lời giải:

\(A=\left|x^2+x+1\right|+\left|x^2+x-12\right|\)

\(A=\left|x^2+x+1\right|+\left|12-x^2-x\right|\)

Áp dụng bđt: \(\left|a\right|+\left|b\right|\ge\left| a+b\right|\) ta có:

\(A=\left|x^2+x+1\right|+\left|12-x^2-x\right|\ge\left|x^2+x+1+12-x^2-x\right|=13\)

Dấu "=" xảy ra khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x+1\ge0\\12-x^2-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x+1\le0\\12-x^2-x\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x\ge-1\\x^2+x\le12\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x\le-1\\x^2+x\ge12\end{matrix}\right.\end{matrix}\right.\)

Vậy \(-1\le x^2+x\le12\) thì \(min_a=13\)