Question

Tìm x, biết:

\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{2016}\)

Mình cần giải cụ thể ko cần kết quả    ~~

 

Answer 1

\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2016}\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2016}\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{1}{2016}\Leftrightarrow\frac{x+1-5}{5\left(x+1\right)}=\frac{1}{2016}\Leftrightarrow\frac{x-4}{5x+5}=\frac{1}{2016}\Rightarrow2016\left(x-4\right)=5x+5\Leftrightarrow2016x-8064=5x+5\)còn lại tự giải

Answer 2

Bài dễ mà !

Answer 3

=> \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2016}\)

=> \(\frac{1}{5}-\frac{1}{x+1}=\frac{1}{2016}\)

=> \(\frac{1}{x+1}=\frac{1}{5}-\frac{1}{2016}=\frac{2011}{10080}\)

=> \(x+1=1:\frac{2011}{10080}=\frac{10080}{2011}\)

=> x= 8069/2011