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!!!
