Question

[Đông Phương Vô Hàn]

Bài 1: Tính giá trị biểu thức 

A= ^{_{​​\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)}}

B= \(\frac{\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+...+\frac{1}{149}}{\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}}\)

C= \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}}\)

D= \(\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.4000}}\)

Bài 2: Tìm x, biết:

a) \(\frac{x+1}{2014}+\frac{x+2}{2013}+...+\frac{x+1007}{1008}=\frac{x+1008}{1007}+\frac{x+1009}{1006}+...+\frac{x+2014}{1}\)

b) \(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-1}{4}\)

Các bạn làm hết giúp mik nha! ^ ^

Answer 1

[Đức Phạm]

Bài 1  :

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\left(\frac{2017}{1}+1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)+1}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\frac{2018}{1}+\frac{2018}{2}+\frac{2018}{3}+....+\frac{2018}{2017}+\frac{2018}{2018}}\)

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

\(=\frac{1}{2018}\)

Answer 2

[Hoang Duc Thinh]

B=\(\frac{\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}}{\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}}\)

\(\)TA CÓ E=\(\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}\)

\(200E=\frac{200}{101.99}+\frac{200}{103.97}+..+\frac{200}{149.51}\)

\(200E=\frac{101+99}{101.99}+\frac{103+97}{103.97}+...+\frac{149+51}{149.51}\)

\(200E=\frac{1}{99}+\frac{1}{101}+\frac{1}{97}+\frac{1}{103}+...+\frac{1}{51}+\frac{1}{149}\)

\(200E=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\)

\(E=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right):200\)\(=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right).\frac{1}{200}\)

\(\Rightarrow B=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}\)/\(\left(\frac{1}{51}+\frac{1}{53}+..+\frac{1}{149}\right).\frac{1}{200}\)

\(\Rightarrow B=\frac{1}{\frac{1}{200}}=200\)

VẬY B=200