\(\dfrac{1}{10}\)

(100/1.101 + 100/2.102 + 100/3.103 +....+100/10.110) . x

= (10/1.11 + 10/2.12 + 10/100.110 )10

=>(1+1/2+1/3+...+1/10-1/101-...-1/110)x

=(1+1/2+1/3+...+1/10+1/11+...+1/100-1/11-...-1/100-1/101-...-1/110)10 =>(1+1/2+1/3+...+1/10-1/101-...-1/110)x

=(1+1/2+1/3+...+1/10-1/101-...-1/110)10 =>x=10

⇒(1−1101 +12 −1102 +13 −1103 +...+110 −1110 ).x=10.(1−111 +12 −112 +...+1100 −1110 )

⇒((1+12 +13 +...+110 )−(1101 +1102 +...+1110 )).x=10.((1+12 +..+110 +111 +112 +...+1100 )−(111 +112 +...+1110 ))

Chu Bạch Dương Linh bạn sửa lại cái đề là mk làm đc.

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

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

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

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

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

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

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

$100E$=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

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