\(\left(\frac{7}{1.2}+\frac{7}{2.3}+\frac{7}{3.4}+...+\frac{7}{2015.2016}\right):\frac{2015}{2016}\)

=\(7\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right):\frac{2015}{2016}\)

=\(7\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\frac{2015}{2016}\)

=\(7\left(\frac{1}{1}-\frac{1}{2016}\right):\frac{2015}{2016}=7.\frac{2015}{2016}:\frac{2015}{2016}=7\)

\(\left(\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+\frac{7}{3\cdot4}+...+\frac{7}{2015\cdot2016}\right):\frac{2015}{2016}\)

\(=\left(7-\frac{7}{2}+\frac{7}{2}-\frac{7}{3}+\frac{7}{3}-\frac{7}{4}+...+\frac{7}{2015}-\frac{7}{2016}\right):\frac{2015}{2016}\)

\(=\left(7-\frac{7}{2016}\right):\frac{2015}{2016}=\frac{2015}{288}:\frac{2015}{2016}=\frac{2015}{288}\cdot\frac{2016}{2015}=\frac{2016}{288}=7\)

\(\frac{2}{1.2}+\frac{2}{2.3}+..........+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)

\(\Rightarrow2\left(\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{x\left(x+1\right)}\right)=\frac{4028}{2015}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{x}-\frac{1}{x+1}=\frac{4028}{2015}:2\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{2014}{2015}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{2014}{2015}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)

\(\Rightarrow x+1=2015\Rightarrow x=2014\)

\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)

\(2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}\right)=1\frac{2013}{2015}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=1\frac{2013}{2015}\div2\)

\(1-\frac{1}{x+1}=\frac{2014}{2015}\)

\(\frac{1}{x+1}=1-\frac{2014}{2015}\)

\(\frac{1}{x+1}=\frac{1}{2015}\)

\(x+1=2015\)

\(x=2015-1\)

\(x=2014\)

Ta thấy A gồm có 99 số hạng nên ta nhóm mỗi nhóm 3 số hạng.

Ta có: A = 1 + 5 + 5² + 5³ + 5⁴ + 5⁵ +...+ 5⁹⁷ + 5⁹⁸ + 5⁹⁹

             = (1 + 5 + 5² )+ (5³ + 5⁴ + 5⁵ )+...+( 5⁹⁷ + 5⁹⁸ + 5⁹⁹ )

             =(1 + 5 + 5² )+ 5³(1 + 5 + 5² ) +...+ 5⁹⁷(1 + 5 + 5² )

             = ( 1 + 5 + 5²)(1 + 5³+....+5⁹⁷)

             = 31(1 + 5³+....+5⁹⁷)

Vì có một thừa số là 31 nên A chia hết cho 31.

 P/s Đừng để ý câu trả lời của mình

Mk nghi~ bn ne^n xem La.i de^` nhe'

à, minh ghi thiếu.

=-1

\(\Leftrightarrow x.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1999}-\frac{1}{2000}\right)=1\)

\(\Leftrightarrow x.\left(1-\frac{1}{2000}\right)=1\Leftrightarrow x\cdot\frac{1999}{2000}=1\Leftrightarrow x=\frac{2000}{1999}\)

\(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2006.2007}=\frac{2006}{2007}\)

\(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+\frac{x}{3}-\frac{x}{4}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)

\(x-\frac{x}{2007}=\frac{2006}{2007}\)

\(\frac{2007x}{2007}-\frac{x}{2007}=\frac{2006}{2007}\)

\(2007x-x=2006\)

\(2006x=2006\)

\(x=1\)

theo mình x =1 không biết đúng không

theo suy luận của mình thì x sẽ bằng 1