\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

\(=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

Giải tiếp(ko chép đề)

= 1/1 - 1/2 - 1/3 - 1/4 + 1/2 - 1/3 - 1/4 - 1/5 + ... + 1/27 - 1/28 - 1/29 - 1/30

= 1 - 1/30

= 29/30

ks nha 

Bài giải :(không chép đề)

=1-1/2-1/3-1/4-1/5+1/2-1/3-1/4-1/5+........+1/27-1/28-1/29-1/30

=1-1/30

=29/30

Vậy số cần tìm là:29/30 Suy ra Y=29/30

\(\frac{1}{1.2.3.4}=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}\right)\)
\(\frac{1}{2.3.4.5}=\frac{1}{3}\left(\frac{1}{2.3.4}-\frac{1}{3.4.5}\right)\)
.........
\(\frac{1}{27.28.29.30}=\frac{1}{3}\left(\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(y=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}\right)+\frac{1}{3}\left(\frac{1}{2.3.4}-\frac{1}{3.4.5}\right)+...+\frac{1}{3}\left(\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(y=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(y=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
CÒN LẠI TỰ LÀM NỐT, MỎI TAY QUÁ RÙI =))

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

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

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

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

\(\Leftrightarrow x\approx0,0648\)

a)\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{100.101}\)=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+....+\(\frac{1}{100}\)-\(\frac{1}{101}\)=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)

b)\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+....+\(\frac{1}{28.29.30}\)=\(\frac{868}{3480}\)=\(\frac{217}{870}\)

c)\(\frac{1}{1.2.3.4}\)+\(\frac{1}{2.3.4.5}\)+....+\(\frac{1}{27.28.29.30}\)=\(\frac{24354}{438480}\)=\(\frac{451}{8120}\)

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)

=> \(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{27.28.29.30}\)

=> \(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

=> \(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}=\frac{14.29.10-1}{28.29.30}=\frac{4059}{28.29.30}\)

=> \(A=\frac{4059}{28.29.30}:3=\frac{1353}{28.29.30}=\frac{451}{28.29.10}\)

=> \(A=\frac{451}{8120}\)

\(\Rightarrow\left(\frac{1}{1}-\frac{1}{30}\right)x=-3\)

\(\Rightarrow\frac{29}{30}x=-3\)

\(\Rightarrow x=\left(-\frac{29}{90}\right)\)

tính trog ngoặc trc nè :

\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

=\(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

=\(\frac{1}{1.2.3}-\frac{1}{28.29.30}\)

=\(\frac{1}{6}-\frac{1}{24360}\)

=\(\frac{1353}{8120}\)

thay vô biểu thức :

\(\frac{1353}{8120}.x=-3\)

x=\(-\frac{8120}{451}\)

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=\(\frac{1}{5}\div3\)=\(\frac{1}{15}\)