= 2/(2.3) + 2/3.4 + 2/4.5 +...+ 2/x(x+1)

= 2 [1/2-1/3+1/3-1/4+...+1/x-1/(x+1)]

=2[1/2-1/(x+1)]= (x-1)/(x+1)

= 2001/2003

==> x=2002

x=2002

Mình Giúp Họ Giải Toán Đầu tiên Mà Họ Lại Làm Ngơ sai bét

=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2

=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)

Với x+2020=0=>x=-2020

Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí 

Vậy x=-2020

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{4034}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2017}\)

\(\Leftrightarrow x+1=2017\)

\(\Leftrightarrow x=2017-1\)

\(\Leftrightarrow x=2016\)

Vậy x = 2016

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2017}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)

\(\Rightarrow2\cdot\frac{x-1}{2\left(x+1\right)}=\frac{2015}{2017}\)

\(\Rightarrow\frac{x-1}{2x+2}=\frac{2015}{4034}\)

\(\Rightarrow4034x-4034=4030x+4030\)

\(\Rightarrow4034x-4030x=8064\)

\(\Rightarrow x=2016\)

câu 1. tìm x nguyên để \(\frac{-35}{6}\)<x<\(\frac{-18}{5}\)

<=> -4,375<x<-3,6

mà x\(\in\)Z nên x={-4}

câu 2. A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)

B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)

Vì \(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)

Vậy B<A

cau3:

\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+.....+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{2007}{2009}\)

2.(\(\frac{1}{6}\)+\(\frac{1}{12}\)+\(\frac{1}{20}\)+.....+\(\frac{1}{x\left(x+1\right)}\))=\(\frac{2007}{2009}\)

2.(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+.....+\(\frac{1}{x\left(x+1\right)}\))=\(\frac{2007}{2009}\)

2.(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{x}\)-\(\frac{1}{x+1}\))=\(\frac{2007}{2009}\)

2.(\(\frac{1}{2}\)-\(\frac{1}{x+1}\))=\(\frac{2007}{2009}\)

\(\frac{1}{2}\)-\(\frac{1}{x+1}\)=\(\frac{2007}{4018}\)

\(\frac{1}{x+1}\)=\(\frac{1}{2}\)-\(\frac{2007}{4018}\)

\(\frac{1}{x+1}\)=\(\frac{1}{2009}\)

x+1=2009

x=2009-1

x=2008