\(\Leftrightarrow\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{375}{376}\)

\(\Leftrightarrow1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{375}{376}\)

\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{375}{376}\)

\(\Leftrightarrow\dfrac{1}{x+3}=1-\dfrac{375}{376}=\dfrac{1}{376}\)

\(\Rightarrow x+3=376\)

\(\Rightarrow x=373\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{667}{2002}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{667}{2002}\)

\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{667}{2002}\) 

\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{x+3}\right)=\frac{667}{2002}\) 

                  \(\frac{1}{1}-\frac{1}{x+3}=\frac{667}{2002}:\frac{1}{3}\)

                   \(\frac{1}{1}-\frac{1}{x+3}=\frac{2001}{2002}\) 

                              \(\frac{1}{x+3}=1-\frac{2001}{2002}\) 

                               \(\frac{1}{x+3}=\frac{1}{2002}\) 

                                \(\frac{1}{x}=\frac{1}{2002-3}\) 

                                 \(\frac{1}{x}=\frac{1}{1999}\)

Vậy x = 1999

đặt VT là A ta có:

\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{6}{19}\)

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

\(A=\left(1-\frac{1}{x+3}\right):3\)

thay A vào VT ta đc:\(\left(1-\frac{1}{x+3}\right):3=\frac{6}{19}\)

\(1-\frac{1}{x+3}=\frac{18}{19}\)

\(\frac{1}{x+3}=\frac{1}{19}\)

=>x+3=19

=>x=16

có thể giải cụ thể ra được ko

1/ 1.4+ 1/ 4.7+ 1/ 7.10+....+1/ x.( x+ 3)= 672/ 2017

(3/1.4+3/4.7+3/7.10+...+ 3/x(x+3)).1/3=672/2017

(1/1-1/4+1/4-1/7+1/7-1/10+.....+(x+3)-x/x.(x+3)).1/3=672/2017

(1/1-1/(x+3)).1/3=672/2017

1/1-1/(x+3)= 672/2017:1/3

1/1-1/(x+3) = 2016/2017

1/(x+3)=1/1-2016/2017

1/(x+3)=1/2017

x+3=2017

x= 2017-3

x= 2014

MIK CHẮC CHẮN 100% LÀ ĐÚNG, DẠNG TOÁN NÀY MIK LM NHIỀU R

HOK TỐT 

\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{672}{2017}\)

\(\Rightarrow\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{672}{2017}\)

\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{672}{2017}\)

\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{x+3}\right)=\frac{672}{2017}\Rightarrow1-\frac{1}{x+3}=\frac{672}{2017}:\frac{1}{3}\)

\(\Rightarrow1-\frac{1}{x+3}=\frac{672}{2017}\cdot3=\frac{2016}{2017}\Rightarrow\frac{1}{x+3}=1-\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+3}=\frac{2017}{2017}-\frac{2016}{2017}\Rightarrow\frac{1}{x+3}=\frac{1}{2017}\)

\(\Rightarrow x+3=2017\Rightarrow x=2017-3\Rightarrow x=2014\)

\(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x.\left(x+3\right)}=\frac{627}{2017}\\ =>\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{627}{2017}\\ =>\frac{1}{3}\left(1-\frac{1}{x+3}\right)=\frac{627}{2017}\\ =>1-\frac{1}{x+3}=\frac{1881}{2017}\)

=>1/x+3=136/2017

.... thôi mỏi tay quá bạn tự làm tiếp đi(ps:bài tớ làm chưa chắc đúng)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)

\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=\dfrac{1}{103}\)

\(\Rightarrow x+3=103\)

\(x=103-3\)

\(x=100\)

Vậy x = 100

x=2009

\(\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}\left(1-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}\times\frac{x+3-1}{x+3}=\frac{6}{19}\)

\(\frac{x+3-1}{x+3}=\frac{6}{19}\div\frac{1}{3}\)

\(\frac{x+2}{x+3}=\frac{18}{19}\)

x = 16

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)

=>\(3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{125}{376}\)

=>\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{375}{376}\)

=>\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{375}{376}\)

=>\(1-\frac{1}{x+3}=\frac{375}{376}\)

=>\(\frac{1}{x+3}=1-\frac{375}{376}\)

=>\(\frac{1}{x+3}=\frac{1}{376}\)

=>x+3=376

=>x=376-3

=>x=373

Vậy x=373

1/1+4 +1/4×7 +1/7×10+.....+1/x×(x+3)=16/49