Theo bài ra, ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2017.2018.2019}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2017.2018.2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2018.2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2018.2019}\right)\)

Giải thích:

\(\frac{2}{1.2.3}=\frac{3}{1.2.3}-\frac{1}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)

\(\frac{2}{2.3.4}=\frac{4}{2.3.4}-\frac{2}{2.3.4}=\frac{1}{1.2}-\frac{1}{3.4}\)

................................................................................

\(\frac{2}{2017.2018.2019}=\frac{2019}{2017.2018.2019}-\frac{2017}{2017.2018.2019}=\frac{1}{2017.2018}-\frac{1}{2018.2019}\)

\(A=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)

\(A=1+\left(1+\frac{2017}{2}\right)+\left(1+\frac{2016}{3}\right)+...+\left(1+\frac{1}{2018}\right)\)

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

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

Ta có: \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)

P=(1/2-1).(1/3-1).(1/4-1)......(1/2017-1).          (1/2018-1)

Ta có:

Số số hạng:(2018-2):1+1=2017( số)

Do 2017 là số lẻ nên,ta có:

P=(-1/2).(-2/3).(-3/4).....(-2015/2016).            (-2016/2017).(-2017/2018)

P=-1/2018

Tinh gia chi bieu thuc                                                                                                                                                                                     2018 : 1/2 + 2018 : 1/3 + 2018 : 1/4 + 2018        

lộn lớp 6

hình như cái này đâu phải toán lớp 5 đâu bạn

nhầm toán lớp 6

12+13×14

= (1 - 1/2).(1 - 1/3).(1 - 1/4) ... (1 - 1/2017)
=1/2 .2/3.3/4.......2016/1017
=1.2.3.4....2016/2.3.4.5...2017
=1.(2.3.4..2016)/(2.3.4..2016).2017
=1/2017( chia cả tử và mẫu cho 2.3.4.2016)

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2016}{2017}.\frac{2017}{2018}\)

\(=\frac{1}{2018}\)

P/S: chúc bạn học tốt

Đặt \(2017=a\)

\(A=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2a+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2\left(a+1\right)\cdot\dfrac{a}{a+1}+\left(\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1-\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\left|a+1-\dfrac{a}{a+1}\right|+\dfrac{a}{a+1}\)

Ta có \(\dfrac{a}{a+1}< 1\Leftrightarrow a+1-\dfrac{a}{a+1}>0\)

\(\Leftrightarrow A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2018\)

2020/2019 x 2019/2018 x 2018/2017 x....................3/2
= 2020/2
= 1010 

\(A=1\dfrac{1}{2019}\times1\dfrac{1}{2018}\times1\dfrac{1}{2017}\times...\times1\dfrac{1}{2}\)

\(=\dfrac{2020}{2019}\times\dfrac{2019}{2018}\times\dfrac{2018}{2017}\times...\times\dfrac{3}{2}\)

\(=\dfrac{2020}{2}\)

\(=1010\)