Question

Tìm x : \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\cdot\left(x+1\right)}\) =\(\frac{2}{9}\) (x thuộc N*)

Answer 1

ta xét VT=\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{x\left(x+1\right)}\right)\)

                =\(2\left(\frac{7-6}{6\cdot7}+\frac{8-7}{7\cdot8}+...+\frac{\left(x+1\right)-x}{x\left(x+1\right)}\right)=2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

                =\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)\)= 2*1/9

=> \(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

<=> \(\frac{1}{x+1}=\frac{1}{18}\)

<=> x+1=18

=> x=17

Answer 2

tớ làm khác nhưng kết quả thì giống