Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

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

\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)

\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)

\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)

\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)

Nguyễn Tiến Đạt

a)\(|3x-5|=|x+2|\)

=> Ta có 2 trường hợp

*) TH1: 3x-5=x+2

=>3x-x=2+5

=>2x=7

=>x=7:2\(\Rightarrow x=\frac{7}{2}\)

*)TH2: -3x+5=x+2

\(\Rightarrow5-3x=x+2\)

\(\Rightarrow5-2=x+3x\)

\(\Rightarrow3=4x\)

\(\Rightarrow x=3:4\Rightarrow x=\frac{3}{4}\)

Vậy \(x\in\left\{\frac{7}{2};\frac{3}{4}\right\}\)

Ta có
\(2017-\left(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+\frac{4}{7}+...+\frac{2017}{2020}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{2}{5}+...+\frac{2017}{2020}\right)\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{2}{5}\right)+...+\left(1-\frac{2017}{2020}\right)\)
\(=\frac{3}{4}+\frac{3}{5}+....+\frac{3}{2020}\)
\(=\frac{3.5}{4.5}+\frac{3.5}{5.5}+\frac{3.5}{6.5}+...+\frac{3.5}{2020.5}\)
\(=3.5\left(\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\right)\)
\(=15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)\)
Thế vào ta có
\(\frac{15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)}{\frac{1}{20}+\frac{1}{25}+...+\frac{1}{10100}}=15\)

Được cập nhật 41 giây trước (17:23)

Ta có :
2017−(14 +25 +36 +47 +...+20172020 )
=(1+1+...+1)−(14 +25 +...+20172020 )
=(1−14 )+(1−25 )+...+(1−20172020 )
=34 +35 +....+32020 
=3.54.5 +3.55.5 +3.56.5 +...+3.52020.5 
=3.5(14.5 +15.5 +16.5 +...+12020.5 )
=15.(1

là luc sắp chết mày yêu ai?

\(\frac{2.2017}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2017}}\)

\(=\frac{4043}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{2017.2018:2}}\)

\(=\frac{4043}{\frac{2}{2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2017.2018}}\)

\(=\frac{4043}{2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)}\)

\(=\frac{4043}{2\left(1+\frac{1}{2}-\frac{1}{2018}\right)}\)

\(=\frac{4043}{2.\frac{1513}{1009}}\)

trog máy tính bấm ra 4034 nưng lại vt 4043 :(( 

sai r bn trẻ ạ :>

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

\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{2016.2018}{2017^2}\)

\(=\frac{2.3^2.4^2.5^2...2016^2.2017.2018}{2^2.3^2.4^2.5^2...2017^2}\)

\(=\frac{2018}{2.2017}=\frac{1009}{2017}\)

lơp 6  ko bt

mk mới học lớp 6 thôi

tk mk nha mk đang âm điểm nè hu hu

Ahuhu giúp zới

mình chưa học đến lớp 8 nên ko thể giúp bạn

=1,25 x 81,25 x 707281,25 / 16,25 x 256,25 x 810000,25

=71833251,95 / 3372891666

=0,0212972307.

k cho mình nha!