Question

\(\frac{2.2017}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+..+2017}}\)Tính giá trị của biểu thức:

 

Answer 1

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

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

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

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

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

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

Answer 2

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

Answer 3

sai r bn trẻ ạ :>

Answer 4

dòng thứ tư trở đi là sai nêm mk làm tiếp nhé

\(=\frac{2.2017}{2-\frac{2}{2018}}=\frac{2.2017}{\frac{2.2017}{2018}}=\frac{\frac{2.2017}{1}}{\frac{2.2017}{2018}}=2018\)

vậy