⇒(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 ))

Ta có:
$(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}$

$\Leftrightarrow \frac{1}{100}\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=\frac{1}{10}\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$

$\Leftrightarrow \left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=10\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$

Đặt $A=\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}$

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

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

$\Rightarrow A=\frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}$

Thay vào phương trình, ta có:

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

$\Leftrightarrow x=10$

\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)

\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+\frac{100}{3\cdot103}+...+\frac{100}{100\cdot110}\right)x=10\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)

\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+....+\frac{1}{10}-\frac{1}{110}\right)x=10\)\(\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)

\(\Rightarrow\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{110}\right)\right]x=10\)\(\left[\left(1+\frac{1}{2}+....+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{110}\right)\right]\)

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

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

\(\Rightarrow x=10\)

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

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

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

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

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Cô giáo cậu giải được đấy sao không hỏi?

\(\dfrac{1}{10}nhébạn\)

E = 1/1.101+1/2.102+...+1/10.110

E = 1/100[100/1.101+100/2.102+...+100/10.110]

E = 1/100[1/1-1/101+1/2-1/102+...+1/10-1/110]

E = 1/100[[1/1+1/2+1/3...+1/10]-[1/101+1/102+...+1/110] - xg cái E

F = 1/1.11+1/2.12+...+1/100.110

F = 1/10[10/1.11+10/2.12+...+10/100.110]

F = 1/10[1/1-1/11+1/2-1/12+...+1/100-1/110]

F = 1/10[[1/1+1/2+...+1/100]-[1/11+1/12...+1/110]]

F = 1/10[[1/1+1/2+...+1/10]-[1/101+1/102+...+1/110]

⇒EF=1100[[11+12+...+110]−[1101+1102+...+1110]]110[[11+12+...+110]−[1101+1102+...+1110]]=110