Bài `1 :`
`a, 12/50 + 8% + 59/100 + 9%`
`= 0,24 + 0,08 + 0,59 + 0,09`
`= 1`
`b, 1/3 + 1/6 + 1/10 + ... + 1/45`
`= 2 × [1/2 × (1/3 + 1/6 + 1/10 + ... + 1/45)]`
`= 2 × (1/6 + 1/12 + 1/20 + ... + 1/90)`
`= 2 × (1/(2 × 3) + 1/(3 × 4) + 1/(4 × 5) + .. + 1/(9 × 10))`
`= 2 × (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10)`
`= 2 × (1/2 - 1/10)`
`= 2 × 2/5`
`= 4/5`
Bài `2 :`
`a,`
- Để `overline{5a1b}` chia hết cho `5` thì `b = 0 ; 5`
- Mà `overline{5a1b}` chia cho `2` dư `1 => b = 5`
- Ta được số : `overline{5a15}`
- Để `overline{5a15}` chia hết cho `3` thì `(5 + a + 5 + 1)` chia hết cho `3`
`=> (11 + a)` chia hết cho `3`
`=> a = 1 ; 4 ; 7`
- Vậy `a = 1 ; 4 ; 7` và `b = 5`
`b,`
- Ta có :
`overline{ab} : b = b` (dư `a`)
`=> (overline{ab} - a) : b = b`
`=> (a × 10 + b - a) : b = b`
`=> [a × (10 - 1) + b] : b = b`
`=> (a × 9 + b) : b = b`
`=> a × 9 + b = b × b`
`=> a × 9 = b × b - b`
- Vì `b ≤ 9 => b × b - b ≤ 9 × 9 - 9 => b × b - 9 ≤ 72`
`=> b = 9` và `a = 8`
Bài `3 :`
`a, (x - 1/2) × 5/2 + 1/2 = 7/4`
`(x - 1/2) × 5/2 = 7/4 - 1/2`
`(x - 1/2) × 5/2 = 5/4`
`x - 1/2 = 5/4 : 5/2`
`x - 1/2 = 2`
`x = 2 + 1/2`
`x = 5/2`
`b, 53,5 = 22 × (x + 1) - 12,5`
`22 × (x + 1) = 53,5 + 12,5`
`22 × (x + 1) = 66`
`x + 1 = 66 : 22`
`x + 1 = 3`
`x = 3 - 1`
`x = 2`
`c, 12 × (x - 6) = 4 × x + 12`
`12 × x - 12 × 6 = 4 × x + 12`
`12 × x - 72 = 4 × x + 12`
`12 × x - (72 + 12) = 4 × x`
`12 × x - 84 = 4 × x`
`84 = 12 × x - 4 × x`
`84 = (12 - 4) × x`
`84 = 8 × x`
`x = 84 : 8`
`x = 10,5`
`d, 2 = (1954 × 0,24 + 76 × 19,54)/(977 × (x - 4))`
`2 = (19,54 × 100 × 0,24 + 76 × 19,54)/(977 × (x - 4))`
`2 = (19,54 × 24 + 76 × 19,54)/(977 × (x - 4))`
`2 = (19,54 × (24 + 76))/(977 × (x - 4))`
`2 = (19,54 × 100)/ (977 × (x - 4))`
`2 = 1954/(977 × (x - 4))`
`=> 977 × (x - 4) = 1954 : 2`
`=> 977 × (x - 4) = 977`
`=> x - 4 = 977 : 977`
`=> x - 4 = 1`
`=> x = 1 + 4`
`=> x = 5`