a) 2/3 – 1/3 (x – 3/2) – 1/2 (2x + 1) = 5
b) 1/3x + 2/5 (x – 1) = 0
a) 2/3 – 1/3 (x – 3/2) – 1/2 (2x + 1) = 5 b) 1/3x + 2/5 (x – 1) = 0
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a) 2/3 – 1/3 (x – 3/2) – 1/2 (2x + 1) = 5
b) 1/3x + 2/5 (x – 1) = 0
Euphemia
a) `2/3 – 1/3 ( x – 3/2 ) – 1/2 ( 2x + 1 ) = 5`
`=> 1/3x – 1/3 . 3/2 – 1/2 . 2x – 1/2 = 2/3 – 5 = -13/3`
`=> 1/3x – 1/2 – 1/2 . 2x – 1/2 = -13/3`
`=> 1/3x – 1/2 . 2x = -13/3 + 1/2 + 1/2 = -10/3`
`=> x . ( 1/3 – 1/2 . 2 ) = -10/3`
`=> x . 4/3 = -10/3`
`=> x = -5/2`
b) `1/3x + 2/5 ( x – 1 ) = 0`
`=> 1/3x + 2/5x – 2/5 = 0`
`=> x . ( 1/3 + 2/5 ) = 2/5`
`=> x . 11/15 = 2/5`
`=> x = 2/5 : 11/15 = 6/11`
Hưng Khoa
a) `x(x – 2) + x – 2 = 0`
`⇔ x(x – 2) + (x – 2) = 0`
`⇔ (x – 2)(x + 1) = 0`
`⇔ x – 2 = 0 hoặc x + 1 = 0`
* `x – 2 = 0`
` x = 2`
* `x + 1 = 0`
` x = -1`
Vậy pt có tập nghiệm: `S = {2; -1}`
`b) 5x(x – 3) – x + 3 = 0`
`⇔ 5x(x – 3) – (x – 3) = 0`
`⇔ (x – 3)(5x – 1) = 0`
`⇔ x – 3 = 0 hoặc 5x – 1 = 0`
* `x – 3 = 0`
`x = 3`
* `5x – 1 = 0`
5x = 1
`x = 1/5`
Vậy pt có tập nghiệm:
`S = {3; 1/5}`