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1 tháng 4 2015

= 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

17 tháng 4 2020

x=2002

7 tháng 4 2023

Ta có:

1/3 + 1/6 + 1/10 + ... + 1/x(x+1):2 = 2001/2003

=> 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 2001/2003

=> 2 [1/6 + 1/12 + 1/20 + ... + 1/x(x+1)] = 2001/2003

=> 2 [1/2x3 + 1/3x4 + 1/4x5 + ... + 1/x+(x+1)] = 2001/2003

=> 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1= 2001/2003 : 2

=> 1/2 - 1/x+1 = 2001/4006

=> 1/x+1 = 1/2 - 2001/4006 = 1/2003

=> x+1 = 2003 = 2002 + 1 

=>x = 2002

26 tháng 10

Câu hỏi 

1/3+1/6+1/10+...+1/x(x+1):2=2001/2003

=Tôi ko biết nên tôi xin hết 

Tron vn

9 tháng 8 2015

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{2001}:2=\frac{1999}{4002}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}=\frac{1}{2001}\)

=> x + 1 = 2001

=> x = 2001 - 1

=> x = 2000

9 tháng 8 2015

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)

   \(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}:\frac{1}{2}\)

  \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)

  \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

      \(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\)

    \(\frac{1}{x+1}=\frac{1}{2001}\)

=> x + 1 = 2001

=> x =    2001 - 1

=> x = 2000 

4 tháng 9 2015

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)

=> x + 1 = 2003

=> x = 2003 - 1

=> x = 2002

18 tháng 4 2023

13+16+110+...+1x(x+1):2=2001200313+16+110+...+1�(�+1):2=20012003

26+212+220+...+2x(x+1)=2001200326+212+220+...+2�(�+1)=20012003

2.(12.3+13.4+14.5+...+1x(x+1))=200120032.(12.3+13.4+14.5+...+1�(�+1))=20012003

12−13+13−14+14−15+...+1x−1x+1=20012003:212−13+13−14+14−15+...+1�−1�+1=20012003:2

12−1x+1=2001400612−1�+1=20014006

=> 1x+1=12−20014006=120031�+1=12−20014006=12003

=> x + 1 = 2003

=> x = 2003 - 1

=> x = 2002