\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{2021}\)

<=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)

<=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)

<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)

<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2042}\)

<=> \(\frac{1}{x+1}=\frac{1}{2021}\)

<=> x + 1 = 2021 

<=> x = 2020

Có phải là bình 6a3 học trường THCS Nguyễn Trãi đúng không 

Bài 2: 

Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)

\(\Leftrightarrow16x+40=90+30\)

\(\Leftrightarrow16x=80\)

hay x=5

Bài 1 :

[( 35 - 5 ) : 3 ]³ + 3

= [30 : 3]³ + 3

= 10³ + 3

= 1000 + 3

= 1003

Đây nha bạn!!!

Chúc bạn học tốt!!!

Câu này dễ quá

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2018}{2019}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{4038}\)

\(\Rightarrow x+1=4038\)

\(\Rightarrow x=4037\)

Vậy \(x=4037\)

\(\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)

\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)

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

\(\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)

\(\frac{1}{x+1}=\frac{1}{4038}\)

\(x=4037\)

Khuyển Dạ Xoa sai dấu rùi bạn 

Tổng của dãy số trên bằng 2041210

tổng là 2043231

Số số hạng của dãy đó là : ( 2021 - 1 ) : 1 + 1 = 2021 ( chữ số )

Tổng của các số hạng đó là : ( 2021 + 1 ) x 2021 : 2 = 2043231

Đáp số 2043231

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x-1\right)}=\)\(\frac{2017}{2019}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x-1\right)}=\frac{2017}{2019}\)

\(2\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right]=\frac{2017}{2019}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\)\(\frac{2017}{2019}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{2019}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\)

\(\frac{1}{x+1}=\frac{1}{2019}\)

x + 1 =2019

     x  = 2019-1 =2018

                       Vậy x = 2018

   \(2\left(\frac{1}{3}.\frac{1}{2}+\frac{1}{6}.\frac{1}{2}+\frac{1}{10}.\frac{1}{2}+....+\frac{2}{x\left(x+1\right)}.\frac{1}{2}\right)=\frac{2017}{2019}\)

=>\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)\)\(=\frac{2017}{2019}\)

=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

=> \(2[\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+....+\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+1}]=\frac{2017}{2019}\)

=>\(2\left(\frac{1}{2}+0+0+....+0-\frac{1}{x-1}\right)=\frac{2017}{2019}\)

=>\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

=>\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{4038}\)

=>\(\frac{1}{x+1}=\frac{1}{2019}\)

=> x+1=2019

=>x=2018

\(\Leftrightarrow1+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{x\left(x+1\right)}=1+\dfrac{2019}{2021}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2019}{2021}\)

\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2019}{2021}\)

\(\Leftrightarrow\dfrac{2}{x+1}=1-\dfrac{2019}{2021}\)

\(\Leftrightarrow\dfrac{2}{x+1}=\dfrac{2}{2021}\)

\(\Leftrightarrow x+1=2021\)

\(\Leftrightarrow x=2020\)