\(\Rightarrow\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right).x=10.\left(\frac{10}{1.10}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)

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

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

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

Nhân 100 vào 2 vế ta được :

(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

Hay thì like nha ! hj hj

bài này bạn lấy trong sách nào vậy

⇒(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:

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

$\Leftrightarrow x=10$

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

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

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

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

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

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

\(\Rightarrow x=10\)

ta gọi phần trong ngoặc là A thì ta có

A nhân x = A

     x= A-A

    x=1

Đặt C = \(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\)

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

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

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

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

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

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

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

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

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

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

Khi đó \(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right)x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)

<=> C.x = B

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

=> \(x=10\)

Vậy 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

gõ xong mém xỉu. :)

làm biếng gõ quá

Cô giáo cậu giải được đấy sao không hỏi?

khi ko mún tích thì tích 1 tích

khi mún tích thì tích 50 tích