\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)

\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)

\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(=>2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

\(2.\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

=> \(1-\frac{1}{x+1}=\frac{4008}{2005}:2=\frac{2004}{2005}\)

\(\frac{1}{x+1}=1-\frac{2004}{2005}=\frac{1}{2005}\)

=>x+1=2005

=>x=2004

1/3 + 1/6 + 1/10 +...+ 2/x(x+1) = 2014/2015

Đ/A là 2004

chúc đồng chí Chế Minh Hải học tốt

Ta có : 

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+1-\frac{2}{x+1}=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=2-\frac{2003}{2005}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2007}{2005}\)

\(\Leftrightarrow\)\(x+1=2:\frac{2007}{2005}\)

\(\Leftrightarrow\)\(x+1=\frac{4010}{2007}\)

\(\Leftrightarrow\)\(x=\frac{4010}{2007}-1\)

\(\Leftrightarrow\)\(x=\frac{2003}{2007}\)

Vậy \(x=\frac{2003}{2007}\)

Chúc bạn học tốt ~ 

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2003}{2005}\)

\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}=\frac{1}{2005}\)

\(\Rightarrow x+1=2005\Rightarrow x=2004\)

câu 2 :

 \(\Leftrightarrow\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}-\frac{x+4}{2005}-\frac{x+5}{2004}-\frac{x+6}{2003}\)=0

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x-2009}{2003}\)=0

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)

\(\Rightarrow x+2009=0\)

\(\Rightarrow x=-2009\)

1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x + 1) = 4007/2004

2/2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1) = 4007/2004

2 × (1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x + 1)) = 4007/2004

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1 = 4007/2004 : 2

1 - 1/x + 1 = 4007/2004 × 1/2

x/x + 1 = 4007/4008

=> x = 4007

a)\(\left(\frac{2}{3}x-\frac{4}{9}\right).\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)

\(\frac{2}{3}x-\frac{4}{9}=0\)hoặc\(\frac{1}{2}+\frac{-3}{7}:x=0\)

\(\frac{2}{3}x=\frac{4}{9}\)hoặc\(-\frac{3}{7}:x=-\frac{1}{2}\)

\(x=\frac{4}{9}:\frac{2}{3}\)hoặc\(x=-\frac{3}{7}:\frac{-1}{2}\)

\(x=\frac{2}{3}\)hoặc\(x=\frac{6}{7}\)

Bài 2:

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)

\(=\frac{1}{2004}\)

Bài 2

=1/2 x 2/3 ... x 2003/2004

=1/2004

2.

1/2004

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