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

\(\frac{45}{188}\)

=\(\frac{31}{94}\)theo mình mới giải là vậy còn nếu không bạn có thể trình bày cách giải cho mình luôn được không?

Ta thấy: 1/1-1/4 = 3/4 = 3.(1/1.4)

           1/4-1/7 = 3/28 = 3.(1/4.7)

A = 3(1/1-1/4+1/4-1/7+...+1/97-1/100)

A = 3.(1-1/100)

A = 3.(99/100)

A = 297/100

\(A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=\frac{1}{3}.\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{3}.\frac{99}{100}\)

\(A=\frac{33}{100}\)

c)1*(1/2-1/3+1/3-1/4+.....+1/91-1/94)

1/2-1/94 ban tu tinh nhe

d)1*(1/1-1/4+1/4-1/7+......+1/91-1/94)

1-1/94 ban tu tinh nhe 

tk nha

a) \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)

\(\Leftrightarrow\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\Leftrightarrow\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)

b) \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)

\(\Leftrightarrow\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{n+3-n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)

c,d dễ bn tách ra rồi trừ đi

\(A.\)\(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n}+\frac{-1}{n+1}=\frac{n+1}{n\left(n+1\right)}+\frac{-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)

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

\(C.\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(D.\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}=\frac{1}{1}-\frac{1}{94}=\frac{93}{94}\)

\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)

Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)

\(B=\frac{1}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)

\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)

A = \(\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

A = 1 + \(\frac{1}{4}\) - \(\frac{1}{4}\) + \(\frac{1}{7}\) - \(\frac{1}{7}\) + \(\frac{1}{10}\) -....- \(\frac{1}{2011}\) + \(\frac{1}{2014}\)

A = 1 + \(\frac{1}{2014}\) = \(\frac{2015}{2014}\)

Sai đề : \(\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)

Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)

\(B=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)

\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)

Chúc bạn học tốt !!!

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\Rightarrow\frac{99}{100}=\frac{0.33.x}{2009}\)

\(\Rightarrow100.0.33.x=99.2009\)

\(\Rightarrow0x=198891\Rightarrow\)không có GT x thỏa mãn

Bài nhìn vô muốn xỉu rồi ='((

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

\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)

b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần

2 ) 

a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)

\(\Rightarrow x=4020\)

tu ma lam nguoi ta con gap hon min nhieu

Ko bt làm mới đăng chứ