Question

Đề bài:

So sánh:

A=1/1.2+1/2.3+1/3.4+1/4.5+...+1/2018.2019 với 1

Ai nhanh và đúng, mk tk cho!!!

Và m.n tk lại mk nhé

Answer 1

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2018\cdot2019}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(A=1-\frac{1}{2019}=\frac{2018}{2019}\)

Mà \(\frac{2018}{2019}< \frac{2019}{2019}=1\)

\(\Rightarrow A< 1\)

Answer 2

#)Giải :

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(A=1-\frac{1}{2019}\)

\(A=\frac{2018}{2019}\)

Vì \(\frac{2018}{2019}< 1\Rightarrow A< 1\)

      #~Will~be~Pens~#

Answer 3

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(=1-\frac{1}{2019}< 1\)

\(\Leftrightarrow A< 1\)