A= \(|\sqrt{x^2}+\sqrt{1}-9|+|\sqrt{x^2}+\sqrt{1}-12|\)
A=\(|x+1-9|+|x+1-12|\)
A=\(|x-8|+|x-11|\)
TH1: x<0
=> A= (-x)-8 + (-x) -11
A=(-x-x)-(8+11)
A=-2x-19
TH2:x>0
=> A=x-8+x-11
A=(x+x)-(8+11)
A=2x-19
Tương tự x=0 sau đấy cậu KL nhé, phần sau mình lười
Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\):
\(\left|\sqrt{x^2+1}-9\right|+\left|\sqrt{x^2+1}-12\right|\)\(=\left|\sqrt{x^2+1}-9\right|+\left|12-\sqrt{x^2+1}\right|\)
\(\ge\left|\left(\sqrt{x^2+1}-9\right)+\left(12-\sqrt{x^2+1}\right)\right|=3\)
Vậy \(A_{min}=3\Leftrightarrow\left(\sqrt{x^2+1}-9\right)\left(12-\sqrt{x^2+1}\right)\ge0\)
\(TH1:\hept{\begin{cases}\sqrt{x^2+1}-9\ge0\\12-\sqrt{x^2+1}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+1\ge81\\x^2+1\le144\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2\ge80\\x^2\le143\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{80}\le x\le\sqrt{143}\\-\sqrt{80}\ge x\ge-\sqrt{143}\end{cases}}\)
\(TH2:\hept{\begin{cases}\sqrt{x^2+1}-9\le0\\12-\sqrt{x^2+1}\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+1\le81\\x^2+1\ge144\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2\le80\\x^2\ge143\end{cases}}\left(L\right)\)