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

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

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

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

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

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

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

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

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

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

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

\(10x=9x+9\)

\(x=9\)

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

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

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

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

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

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

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

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 : \(E=\frac{1}{2007}\)

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1 

2(\(\frac{3}{4}\)-5x)=\(\frac{4}{5}\)-3x

=> \(\frac{6}{4}-10x=\frac{4}{5}-3x\)

=>\(-10x+3x=\frac{4}{5}-\frac{6}{4}\)

=> \(x=\frac{1}{10}\)

2 .

\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)

=>\(\frac{3}{2}-1+4x=\frac{2}{3}-7x\)

=>\(11x=\frac{1}{6}\)

=>x=\(\frac{1}{66}\)

3.

\(3\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

=>\(\frac{3}{2}-3x+\frac{1}{3}=\frac{7}{6}-x\)

=>\(-2x=\frac{-2}{3}\)

=>\(\frac{1}{3}\)

4. câu 4 ko hiểu bạn ơi