Tìm GTNN của biểu thức A = | x - 2017 | + | x + 2018 |

Bài 1:

Ta có: \(\left|x-2017\right|+\left|x+2018\right|=\left|2017-x\right|+\left| x+2018\right|\)

Áp dụng bất đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có:

\(\left|2017-x\right|+\left|x+2018\right|\ge\left|2017-x+x+2018\right|=4035\)

Dấu "=" sảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}2017-x\ge0\\x+2018\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le2017\\x\ge-2018\end{matrix}\right.\Rightarrow-2018\le x\le2017\)

Vậy.....................

Bài 2:

Ta có:

\(\left\{{}\begin{matrix}\dfrac{1}{2!}=\dfrac{1}{1.2}\\\dfrac{1}{3!}=\dfrac{1}{2.3}\\.....\\\dfrac{1}{2017!}< \dfrac{1}{2016.2017}\end{matrix}\right.\)

\(\Rightarrow1+\dfrac{1}{1!}+\dfrac{1}{2!}+....+\dfrac{1}{2017!}< 1+1+\dfrac{1}{1.2}+...+\dfrac{1}{2016.2017}\)

Ta lại có:

\(1+1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{2016.2017}\)

\(=2+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2016}-\dfrac{1}{2017}\)

\(=2+1-\dfrac{1}{2017}=3-\dfrac{1}{2017}\)

\(\Rightarrow1+1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{2016.2017}< 3\)

Do đó: \(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+....+\dfrac{1}{2017!}< 3\)(đpcm)

Chúc bạn học tốt!!!