Question

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

Help me...e...!!!!!

 

Answer 1

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

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

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

\(\Leftrightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{2017}{2019}\div2\)

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

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

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2019}\)

\(\Leftrightarrow x+1=2019\)

\(\Leftrightarrow x=2019-1\)

\(\Leftrightarrow x=2018\)

Vậy x = 2018 

Answer 2

chị ơi bài này em học từ lớp 6 rồi( ngay đầu vào luôn ) nhung dốt nát lè

Answer 3

Tao dang bai cho thang Dung  nha con Huyen thoi kia

Answer 4

THÌ BÀI NÀY Ở LỚP 6 CON GÌ CHỊ Ý ĐĂNG THÌ CÓ SAO ĐÂU