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

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

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

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

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

<=> \(\frac{1}{x+1}=\frac{1}{2021}\)

<=> x + 1 = 2021 

<=> x = 2020

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

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 \(x=2020\).

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

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

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

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

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

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

\(\Leftrightarrow x+1=2021\)

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

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

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

\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..........+\frac{1}{x\left(x+1\right)}\right]=\frac{2019}{2021}\)

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

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

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

\(\Leftrightarrow\frac{1}{x-1}=\frac{1}{2021}\)

\(\Leftrightarrow x-1=2021\)

\(\Leftrightarrow x=2022\)

Vậy \(x=2022\)

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

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

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

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

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

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

\(\Rightarrow x+1=4038\)

\(\Rightarrow x=4037\)

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 

Điều kiện đã cho \(\Leftrightarrow7\left(x-2019\right)^2+y^2=23\) (*)

Do \(\left(x-2019\right)^2,y^2\ge0\) nên (*) suy ra \(y^2\le23\Leftrightarrow y^2\in\left\{0,1,4,9,16\right\}\)

\(\Leftrightarrow y\in\left\{0,1,2,3,4\right\}\)

Hơn nữa, lại có \(y^2=23-7\left(x-2019\right)^2\). Ta thấy \(VP\) chia 7 dư 2.

\(\Rightarrow y^2\) chia 7 dư 2 \(\Rightarrow y\in\left\{3,4\right\}\)

Xét \(y=3\) \(\Rightarrow7\left(x-2019\right)^2=14\) \(\Leftrightarrow\left(x-2019\right)^2=2\), vô lí.

Xét \(y=4\Rightarrow7\left(x-2019\right)^2=7\) \(\Leftrightarrow\left(x-2019\right)^2=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=2020\\x=2018\end{matrix}\right.\)

Vậy \(\left(x,y\right)\in\left\{\left(4;2020\right),\left(4;2018\right)\right\}\) thỏa mãn ycbt.

câu trả lời ngắn gon nhất là .................................................................................................................................................................................................................................................................................................................. tự làm nhé bạn

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