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19 tháng 4 2019

Đáp án cần chọn là: C

1 3 + 1 6 + 1 10 + ... + 1 x ( x + 1 ) : 2 = 2019 2021 2. [ 1 2.3 + 1 3.4 + ... + 1 x ( x + 1 ) ] = 2019 2021 2. ( 1 2 − 1 3 + 1 3 − 1 4 + ... + 1 x − 1 x + 1 ) = 2019 2021 2. ( 1 2 − 1 x + 1 ) = 2019 2021 1 − 2 x + 1 = 2019 2021 2 x + 1 = 1 − 2019 2021 2 x + 1 = 2 2021 x + 1 = 2021 x = 2020

14 tháng 7 2021

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

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

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

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

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

<=> \(\frac{1}{x+1}=\frac{1}{2021}\)

<=> x + 1 = 2021 

<=> x = 2020

16 tháng 7 2021

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

30 tháng 7 2020

Đề bạn thiếu 1 số \(x\) nữa đúng không?

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)

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

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

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2021}\)

\(\Rightarrow x+1=2021\)

\(\Rightarrow x=2020\)

Vậy \(x=2020\).

30 tháng 7 2020

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)

\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)

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

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4042}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2019}{4042}=\frac{1}{2021}\)

\(\Leftrightarrow x+1=2021\)

\(\Leftrightarrow x=2020\left(tm:x\in N\right)\)

Bài 2: 

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

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

\(\Leftrightarrow16x=80\)

hay x=5

5 tháng 10 2021

Bài 1 :

[( 35 - 5 ) : 3 ]3 + 3

= [30 : 3]3 + 3

= 103 + 3

= 1000 + 3

= 1003

Đây nha bạn!!!

Chúc bạn học tốt!!!hihi

22 tháng 4 2019

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

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

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

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

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

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

\(\Rightarrow x+1=4038\)

\(\Rightarrow x=4037\)

Vậy \(x=4037\)

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

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

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

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

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

\(x=4037\)

10 tháng 5 2022

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8 tháng 5 2018

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

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

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

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

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

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

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

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

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

x + 1 =2019

     x  = 2019-1 =2018

                       Vậy x = 2018

8 tháng 5 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

NV
25 tháng 3 2021

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

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

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

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

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

\(\Leftrightarrow x+1=2021\)

\(\Leftrightarrow x=2020\)