\(100E\)\(=\frac{100}{1.101}+\frac{100}{2.102}+..........+\frac{100}{10.110}\)

\(=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+........+\frac{1}{10}-\frac{1}{110}\)

\(10F=\frac{10}{1.11}+\frac{10}{2.12}+......+\frac{10}{100.110}\)

\(=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+......+\frac{1}{100}-\frac{1}{110}\)

\(=1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+....+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-....-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{110}\)

\(=1+\frac{1}{2}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\)\(=100E\)

\(\Rightarrow10F=100E\Rightarrow\frac{E}{F}=\frac{1}{10}\)

$100E$=1001.101+1002.102+..........+10010.110=1.101100​+2.102100​+..........+10.110100​

=1−1101+12−1102+........+110−1110=1−1011​+21​−1021​+........+101​−1101​

10F=101.11+102.12+......+10100.11010F=1.1110​+2.1210​+......+100.11010​

=1−111+12−112+......+1100−1110=1−111​+21​−121​+......+1001​−1101​

=1+12+...+110+111+....+1100−111−112−....−1100−1101−...−1110=1+21​+...+101​+111​+....+1001​−111​−121​−....−1001​−1011​−...−1101​

=1+12+...+110−1101−1102−...−1110=1+21​+...+101​−1011​−1021​−...−1101​$=100E\; lm\; như\; bn\; này\; nha\; bn$

⇒10F=100E⇒EF=110

\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103+...}+\frac{1}{10.110}\)

\(A=\frac{1}{100}(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110})\)

\(A=\frac{1}{100}(\frac{1}{1}-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110})\)

\(A=\frac{1}{100}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\)     ok?

\(B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)

\(B=\frac{1}{10}(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110})\)

\(B=\frac{1}{10}(\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110})\)

\(B=\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{100})-(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}))\)=\(\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))\)

B=10A 

A.x=10A suy ra x=10

gõ xong mém xỉu. :)

làm biếng gõ quá

Cô giáo cậu giải được đấy sao không hỏi?

Sửa đề: 

\(E=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\)

=> \(100.E=\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\)

\(=\frac{101-1}{1.101}+\frac{102-2}{2.102}+\frac{103-3}{3.103}+...+\frac{110-10}{10.110}\)

\(=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{110}\right)\)

\(F=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.110}\)

=> \(10F=\frac{10}{1.11}+\frac{10}{2.12}+\frac{10}{3.13}+...+\frac{10}{100.110}\)

\(=\frac{11-1}{1.11}+\frac{12-2}{2.12}+\frac{13-3}{3.13}+...+\frac{110-100}{100.110}\)

\(=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{110}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)+\left(-\frac{1}{11}+\frac{1}{11}\right)+\left(-\frac{1}{12}+\frac{1}{12}\right)+...+\left(-\frac{1}{100}+\frac{1}{100}\right)\)

\(-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)=100E\)

=> 10 F = 100 E

=> \(\frac{E}{F}=\frac{10}{100}=\frac{1}{10}\)

Câu hỏi của Huỳnh Ngọc Cẩm Tú - Toán lớp 6 - Học toán với OnlineMath

Vì gõ trên Hoc24 khá lâu nên mình gửi hình ảnh cho lẹ

$\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{103.105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{103}-\frac{1}{105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{1}{2}.\left(1-\frac{1}{105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{52}{105}.\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{52}{105}x-\frac{52}{105}=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow x=-\frac{3}{11}$

b) Đặt \(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\)

A\(=\frac{1}{100}\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{1}{3.103}+...+\frac{100}{10.110}\right)\)

A\(=\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)

A\(=\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)Đặt \(B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{10}{100.110}\)

\(B=\frac{1}{10}\left(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)

\(B=\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)

\(B=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right]\)\(=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)\(B=10A\)

\(A.x=10A\)

\(=>x=10\)

thanh niên ngu méo làm được bài này