Question

Tìm x:

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}=\frac{2015}{2016}\)

Answer 1

1-1/x+1=2015/2016

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

=>x+1=2016=>x=2015

Answer 2

mình không ghi lại đề nha:

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

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

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

=>x=

Đến đó bạn tự giải tiếp ha

Answer 3

=>(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/x-1/x+1)=2015/2016
=>1/1-1/x+1=2015/2016
=>x/x+1=2015/2016
=>x=2015

Answer 4

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

=\(\frac{1}{1}-\frac{1}{x+1}=\frac{2015}{2016}\)(rút gọn số đối)

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

=> x=2015