a) Đặt A=\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{98\cdot99\cdot100}\)

\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+....+\frac{2}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+.....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)

2A=\(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}=\frac{4949}{9900}\) =>A=\(\frac{4949}{9900}\div2=\frac{4949}{19800}\)

Đặt B=\(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29\cdot30}\)

=>3B=\(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+....+\frac{3}{27\cdot28\cdot29\cdot30}\)

3B=\(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\)

3B=\(\frac{1}{1\cdot2\cdot3}-\frac{1}{28\cdot29\cdot30}=\frac{1353}{8120}\)

=>B=\(\frac{1353}{8120}\div3=\frac{451}{8120}\)

Ta có : A-3x=B=>3x=A-B=\(\frac{4949}{19800}\)-\(\frac{451}{8120}\)\(\approx\frac{1}{5}\)=>x=\(\frac{1}{5}\div3\)=\(\frac{1}{15}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

\(\Rightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)

\(\Rightarrow3x=\frac{4949}{19800}-\frac{451}{8120}\)

\(\Rightarrow x=\left(\frac{4949}{19800}-\frac{451}{8120}\right):3\)

a)Ta có   \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

=)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

=)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=)   \(x+3=305\)=) \(x=302\)

\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)

\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(\Leftrightarrow x=308-3\)

\(\Leftrightarrow x=305\)

Vậy \(x=305\)

Tìm x biết :

 a)  \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)(1)

Gọi B = \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\)

     \(\Rightarrow3A=3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)\)  

     \(\Rightarrow3A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}\)

     \(\Rightarrow3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\)

     \(\Rightarrow3A=\frac{1}{5}-\frac{1}{x+3}\)

      \(\Rightarrow A=\left(\frac{1}{5}-\frac{1}{x+3}\right):3\)(2)

Thay (2) vào (1) , ta có : \(\left(\frac{1}{5}-\frac{1}{x+3}\right):3=\frac{101}{1540}\)

                                 \(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

                                 \(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)

                                 \(\Rightarrow x+3=308\)

                                 \(\Rightarrow x=305\)

                                  Vậy x=305

Tương tự như câu b) nha 

Chúc bạn học tốt!! nhớ mik nha

.....

<=>\(\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

<=>\(\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{202}{1540}\)

<=>\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

<=>\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{202}{1540}:\frac{2}{3}=\frac{303}{1540}\)

<=>\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

<=> x+3=308

<=> x=305

Mình không viết lại đề bài nha

a) \(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\Rightarrow x=305\)

Tìm x,y thuộc Z biết:

a, \(2^{x+y}=2^x+2^y\)

b, \(x+y=x.y=x:y\left(y\ne0\right)\)

Làm nhanh giùm mình nhé!!!!!

\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+..+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{202}{1540}:\frac{2}{3}=\frac{303}{1540}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

\(\Rightarrow x+3=308\Rightarrow x=305\)

Vạy x = 305