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

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

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

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

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

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

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

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

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

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

\(10x=9x+9\)

\(x=9\)

\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{\left(3x-1\right)\left(3x+3\right)}=\dfrac{3}{10}\) \(\Rightarrow\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{15}+...+\dfrac{1}{\left(3x-1\right)}-\dfrac{1}{\left(3x+3\right)}=\dfrac{3}{10}\)\(\Rightarrow\dfrac{1}{3}-0-0-...-0-\dfrac{1}{\left(3x+3\right)}=\dfrac{3}{10}\)(cộng số đối)

\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{\left(3x+3\right)}=\dfrac{3}{10}\)

\(\Rightarrow\dfrac{1}{\left(3x+3\right)}=\dfrac{1}{3}-\dfrac{3}{10}\)

\(\Rightarrow\dfrac{1}{\left(3x+3\right)}=\dfrac{1}{30}\)

\(\Rightarrow3x+3=30\)

\(\Rightarrow x=\left(30-3\right)+3=9\)

Vậy x=9

Viết vậy đúng đó em

A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)

= 5/4 . [4/(3.7) + 4/(7.11) + 4/(11.15) + ... + 4/(2019.2023)]

= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)

= 5/4 . (1/3 - 1/2023)

= 5/4 . 2020/6069

= 2525/6069

câu 3:x=9

câu 6:x=1

y=2

a=4/3.7 +4/7.11+4/11.15 +.....+4/107/111

=1/3-1/7+1/7-1/11+1/11-1/15+......+1/107-1/111

=1/3-1/111

=12/37

12/37nhe

\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{x\left(x+4\right)}=\frac{43}{552}\)

\(\Leftrightarrow\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{x}-\frac{1}{x+4}\right)=\frac{43}{552}\)

\(\Leftrightarrow\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{x+4}\right)=\frac{43}{552}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+4}=\frac{43}{552}\div\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+4}=\frac{43}{138}\Leftrightarrow\frac{1}{x+4}=\frac{1}{3}-\frac{43}{138}\)

\(\Leftrightarrow\frac{1}{x+4}=\frac{1}{46}\Leftrightarrow x+4=46\Rightarrow x=46-4=42\)

Vậy x = 42 

  \(s=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{x\left(x+4\right)}=\)\(\frac{43}{552}\)

\(\Rightarrow S=\frac{4}{4}\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{x\left(x+4\right)}\right)=\frac{43}{552}\)

\(\Rightarrow S=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{x\left(x+4\right)}\right)=\frac{43}{552}\)

\(\Rightarrow S=\frac{1}{4}\left(\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+...+\frac{4}{x}-\frac{4}{x+4}\right)=\frac{43}{552}\)

\(\Rightarrow S=\frac{1}{4}\left(\frac{4}{3}-\frac{4}{x+4}\right)=\frac{43}{552}\)

\(\Rightarrow\frac{4}{3}-\frac{4}{x+4}=\frac{43}{552}:\frac{1}{4}\)

\(\frac{\Rightarrow4}{3}-\frac{4}{x+4}=\frac{43}{138}\)

\(\frac{\Rightarrow4}{x+4}=\frac{4}{3}-\frac{43}{138}=\frac{47}{46}\)

\(\Rightarrow x+4=4:\frac{47}{46}=\frac{184}{47}\)

\(\Rightarrow x=\frac{184}{47}-4=\frac{-4}{47}\)

Đặt A=đã cho

=>\(4A=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{x\cdot\left(x+4\right)}\)

=>\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+4}=\frac{1}{3}-\frac{1}{n+4}\)

=>\(4\cdot\frac{43}{552}=\frac{1}{3}-\frac{1}{x+4}\)

=>\(\frac{43}{138}=\frac{46}{138}-\frac{1}{x+4}\left(1\right)\)

Từ đt (1),ta có thể suy ra 1/x+4=3/138

=>3*(x+4)=138

=>x+4=46

=>x=42

Vậy x=42