Ta có: \(A=1\cdot99+2\cdot98+\cdots+99\cdot1\)

\(=2\left(1\cdot99+2\cdot98+\cdots+49\cdot51\right)+50\cdot50\)

\(=2\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+49\left(100-49\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\cdot49\cdot\frac{50}{2}-\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\cdot49\cdot25-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\cdot49\cdot25-49\cdot25\cdot33\right\rbrack+2500=2\cdot25\cdot49\left(100-33\right)+2500\)

\(=50\cdot49\cdot67+2500=166650\)

Sửa đề: \(B=1\cdot101+2\cdot102+\cdots+9\cdot109\)

\(=1\left(100+1\right)+2\left(100+2\right)+\cdots+9\left(100+9\right)\)

=100(1+2+...+9)+(\(1^2+2^2+\cdots+9^2\) )

\(=100\cdot9\cdot\frac{10}{2}+\frac{9\left(9+1\right)\left(2\cdot9+1\right)}{6}\)

\(=900\cdot5+\frac{9\cdot10\cdot19}{6}=4500+3\cdot5\cdot19=4500+15\cdot19\)

=4500+285

=4785

A+B

=166650+4785

=171435

A+B = (1.99+2.98+3.97+...+99.1)+(1.101+2.102+3.103+...+99.199)

A+B = (1.99+1.101)+(2.98+2.102)+(3.97+3.103)+...+(99.1+99.199)

A+B = 1(99+101) + 2(98+102) + 3(97.103)+...+99(1+199)

A+B = 1.200 + 2.200 + 3.200 +...+ 99.200

A+B = 200.(1+2+3+...+200)

A+B = 200.4950

A+B = 990000

A + B = ( 1 . 99 + 2 . 98 + 3 . 97 + ... + 99 . 1 ) + ( 1 . 101 + 2 . 102 + 3 . 103 + ... + 99 . 199 )

A + B = 99 . ( 1 + 199 ) + 98 . ( 2 + 198 ) + 97 . ( 3 + 197 ) + ... + 2 . ( 102 + 98 ) + 1 . ( 99 + 101 ) 

A + B = 99 . 200 + 98 . 200 + 97 . 200 + ... + 2 . 200 + 1 . 200

A + B = ( 99 + 98 + 97 + ... + 2 + 1 ) . 200

A + B = 4950 . 200

A + B = 990000

A+B=(1.99+2.98+...+99.1)+(1.101+2.102+...+99.199)

=(1.99+1.101)+(2.98+2.102)+...+(99.1+99.199)

=1.(99+101)+2.(98+102)+...+99(1+199)

=200+2.200+...+99.200

=200.(1+2+3+4+...+99)

=200.4950

=.....

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:

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

$\Leftrightarrow x=10$

(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