a)\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)

=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}+\frac{1}{94}-\frac{1}{97}\)(giản ước các phân số giống nhau)

=\(\frac{1}{1}-\frac{1}{97}\)

=\(\frac{96}{97}\)

a)    gọi \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.11}+...+\frac{2}{94.97}\)

               \(\Rightarrow\frac{3}{2}A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}\)

                     \(\frac{3}{2}A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\)(rút gọn các phân số giống nhau)

                      \(\frac{3}{2}A=\frac{1}{1}-\frac{1}{97}\)

                       \(\frac{3}{2}A=\frac{96}{97}\left(1\right)\)

                       từ \(\left(1\right)\Leftrightarrow A=\frac{96}{97}\div\frac{3}{2}=\frac{64}{97}\)

b)\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{2011}\right)\)

    \(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}......\frac{2010}{2011}\)

 \(=\frac{6.7.8.9.....2010}{7.8.9......2011}\)(rút gọn các số giống nhau)

\(=\frac{6}{2011}\)

Câu a Nguyễn Triệu Yến Nhi đúng

Cậu b bao quynh Cao đúng

\(B=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\frac{102}{103}\)

\(B=\frac{34}{103}\)

Bài 3: đổi ra phân số rồi tính, đổi:\(1,5=\frac{15}{10};2,5=\frac{25}{10};1\frac{3}{4}=\frac{7}{12}\)(cái này ko giải dùm, đổi ra như thek rồi tính nha)

\(B=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\frac{102}{103}\)

\(=\frac{1}{1}.\frac{34}{103}=\frac{34}{103}\)

Thằng an mất dạy :))) 
#Lê_Linh

bạn ơi như là cô giáo cho đề sai rồi kết quả phải là \(\frac{375}{376}\)thì mới giải được

Ta có:

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

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

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{125}{376}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{x+3}=\frac{125}{376}:\frac{1}{3}=\frac{375}{376}\)

\(\Rightarrow\frac{1}{x+3}=1-\frac{375}{376}=\frac{1}{376}\Leftrightarrow x+3=376\Leftrightarrow x=373\)

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

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

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

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

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

\(\frac{x+2}{x+3}=\frac{375}{376}\)

=> x + 2 = 375

=> x = 375 - 2

=> x = 373

\(\frac{1}{1.3}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{2018}{6057}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\right)=\frac{2018}{6057}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{2018}{6057}.3\)

\(\Rightarrow1-\frac{1}{n+3}=\frac{2018}{2019}\)

\(\Rightarrow\frac{1}{n+3}=1-\frac{2018}{2019}\)

\(\Rightarrow\frac{1}{n+3}=\frac{1}{2019}\)

\(\Rightarrow n+3=2019\)

\(\Rightarrow n=2016\)

Vậy n = 2016

\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}=\frac{18}{19}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{18}{19}\)

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

...............

đặt VT là A ta có:

\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{6}{19}\)

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

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

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

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

\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{6}{19}\)

\(\Rightarrow\frac{1}{3}\cdot\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{6}{19}\)

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

\(\Rightarrow1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)

\(\Rightarrow1-\frac{1}{x+3}=\frac{6}{19}\cdot3=\frac{2}{19}\)

\(\Rightarrow\frac{1}{x+3}=1-\frac{2}{19}=\frac{17}{19}\)

=>17*(x+3)=19

x+3=19/17

x=19/17-3

x=-32/17

nếu thấy đúng thì k cho mh nhaThái Lâm Hoàng

1/3.(1-1/4+1/4-1/7+......+1/x-1/(x+3)=6/19

1/3.(1-1/x+3)=6/19

1-1/x+3=6/19:1/3

1-1/x+3=18/19

1/x+3=1-18/19

1/x+3=1/19

=> x+3=19

=>x=19-3

x=16    

Đặt biểu thức là A, ta có:

3A=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{x\left(x+3\right)}\)

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

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

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

=>\(\frac{1}{3}-\frac{3}{x+3}\)  =\(\frac{6}{19}\) =>x=168

1/3.(1-1/4+1/4-1/7+......+1/x-1/(x+3)=6/19

1/3.(1-1/x+3)=6/19

1-1/x+3=6/19:1/3

1-1/x+3=18/19

1/x+3=1-18/19

1/x+3=1/19

=> x+3=19

=>x=19-3

x=16    

Pk bt tổng này bằng bao nhiêu ms tính đc chứ

3. ( 1/1.4 +1/4.7 +1/7.10 +...+ 1/x.(x+3)

3/1.4 +1/4.7+1/7.10 + ...+ 3/ x . (x+3)

1/1 - 1/4 + 1/4 - 1/6 + 1/7 - 1/10 + ...+ 1/x-1/x+3

1/1 - 1/x+3

x+3/x+3 - 1/x+3

x+2/x+3

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

\(3A=\frac{3}{1.4}+\frac{4}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\)

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

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

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

Cái này mình rút  gọn rồi đó