\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)(Vì 3x + 3 lớn hơn 3x - 1 là 4 đơn vị)

\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{x+1-1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{x}{3x+3}=\frac{3}{10}\)

\(\Rightarrow10x=3.\left(3x+3\right)\)

\(\Rightarrow10x=9x+9\)

\(\Rightarrow x=9\)

Vậy...

thanks

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-....-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)

Nên 3x + 3 = 30

3x = 30 - 3 = 27

x = 27 : 3 = 9 

4( 1/3.7 + 1/7.11 +...+1/(3x-1)(3x+3) = 3/10

4( 1/3 - 1/7 + 1/7 - 1/11 +...+ 1/(3x-1) - 1/(3x+3) =3/10

4(1 -  1/(3x+3))= 3/10

phần còn lại tự giải nghe x=-71/111

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}=\frac{10}{30}-\frac{9}{30}=\frac{1}{30}\)

\(\Rightarrow\left(3x+3\right).1=1.30\Rightarrow3x+3=30\Rightarrow3x=27\Rightarrow x=9\)

\(\frac{1}{3}-\frac{1}{3x+3}=\frac{3x+3}{9x+9}-\frac{3}{9x+9}=\frac{3x}{9x+9}=\frac{3}{10}\)

\(\frac{x}{3x+3}=\frac{3}{10}\)

\(10x=9x+9\)

\(x=9\)

2+12345678-5=

câu 3:x=9

câu 6:x=1

y=2

\(E=\frac{\frac{4}{3\cdot7}-\frac{4}{11.15}}{1-\frac{3}{7}-\frac{3}{11}+\frac{1}{5}}-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}\right)\)

\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{11}+\frac{1}{15}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2006}-\frac{1}{2007}\right)\)

\(=\frac{\frac{64}{385}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{2007}\right)\)

\(=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{2007}\right)=\frac{1}{2007}\)

Vậy : \(E=\frac{1}{2007}\)