`b)` `(1/2+1/3+1/4+...+1/9+1/10)x=1/9+2/8+3/7+....+8/2+9/1` Tìm x ạ <3

`Ta có: 1/9 + 2/8 + 3/7 + ... + 8/2 + 9/1`

       `= 1/9 + 2/8 + 3/7 + ... + 8/2 + 1 × 9`

       `= ( 1/9 + 1 ) + ( 2/8 + 1 ) + ( 3/7 + 1 ) + ... + ( 8/2 + 1 ) + 1`

       `= 10/9 + 10/8 + 10/7 + ... + 10/2 + 10/10`

       `= 10 × ( 1/10 + 1/9 + 1/8 + 1/7 + ... + 1/2 )`

`⇒ ( 1/2 + 1/3 + 1/4 + ... + 1/9 + 1/10 ) × x = ( 1/10 + 1/9 + 1/8 + ... + 1/3 + 1/2 ) × 10`

`⇒ x = 10`

`Vậy x = 10`