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

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

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

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

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

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

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

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

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

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

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

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

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

\(x+1=2015\)

\(x=2015-1\)

\(x=2014\)

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

1/1.2 +1/2.3 +...+ 1/x(x+1) = 2015/2016

<=> 1-1/2 + 1/2 - 1/3 + ... + 1/x - 1/x+1 = 2015/2016

<=> 1 - 1/x+1 = 2015/2016

<=> 1/x+1 = 1/2016

<=> x + 1 = 2016

<=> x = 2015

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

\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\)

 \(\Leftrightarrow\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)

\(\Leftrightarrow x+1=2016\Rightarrow x=2015\)

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

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(1-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)

\(x=2016-1=2015\)

            Đáp số: 2015

\(\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2013}{2015}\)

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

=>\(1-\dfrac{2}{x+1}=\dfrac{2013}{2015}\)

=>\(\dfrac{2}{x+1}=\dfrac{2}{2015}\)

=>x+1=2015

=>x=2014

2/1.2+2/2.3+2/3.4+...+2/x(x+1)=4028/2015

2(1/1.2+1/2.3+1/3.4+...+1/x(x+1))=4028/2015

2(1/1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/x+1)=4028/2015

2(1-1/x+1)=4028/2015

1-1/x+1=2014/2015

(x+1-1)/x+1=2014/2015

x/x+1=2014/2015

(x+1).2014=2015x

2014x-2015x=-2014

-x=-2014

x=2014

a) A = (1.1)/(1.2) x (2.2)/(2.3) x ... (99.99)/(99.100) = 1/2 . 2/3 . 3/4. ..99/100 = 1/100 

Tk cho em nha

Ta có: 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3

=> 1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2/3

=>1-1/x+1=2/3

=>1/x+1=1/3

=>3=x+1

=>x=2

Ta có\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)

=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{3}\)

=>\(1-\frac{1}{x+1}=\frac{2}{3}\)

=>\(\frac{1}{x+1}=1-\frac{2}{3}\)

=>\(\frac{1}{x+1}=\frac{1}{3}\)

=>\(x+1=3\)

=>\(x=2\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)

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

\(\Rightarrow1-\frac{1}{x-1}=\frac{2}{3}\)

\(\Rightarrow\frac{1}{3}=\frac{1}{x-1}\)

\(\Rightarrow x=3+1=4\)

khó lắm:))