\(A=\sqrt{\left(1989-x\right)^2}+\sqrt{\left(1990-x\right)^2}=\left|x-1989\right|+\left|1990-x\right|\)
Áp dụng bất đẳng thức : \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\). Dấu "=" xảy ra khi a,b cùng dấu.
\(A=\left|x-1989\right|+\left|1990-x\right|\ge\left|x-1989+1990-x\right|=1\)
\(\Rightarrow A\ge1\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1989\ge0\\1990-x\ge0\end{cases}}\Leftrightarrow1989\le x\le1990\)
Vậy Min A = 1 \(\Leftrightarrow1989\le x\le1990\)